NO Termination w.r.t. Q proof of ../tpdb/TRS/higher-order/Bird/Hamming.trs
Termination w.r.t. Q of the following Term Rewriting System could be disproven:

Q restricted rewrite system:
The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

Q restricted rewrite system:
The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming


Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), app(app(cons, y), ys))
LIST2 → APP(s, 0)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys))))
LIST3 → HAMMING
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(lt, x)
HAMMING → APP(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))
HAMMING → LIST3
LIST3 → APP(s, 0)
LIST1 → HAMMING
APP(app(mult, app(s, x)), y) → APP(app(plus, y), app(app(mult, x), y))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys)))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(eq, x)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))
LIST1 → APP(map, app(mult, app(s, app(s, 0))))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(mult, app(s, x)), y) → APP(plus, y)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(lt, x), y)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, app(app(cons, x), xs)), ys)
HAMMING → APP(merge, list2)
APP(app(mult, app(s, x)), y) → APP(mult, x)
LIST3 → APP(s, app(s, app(s, app(s, app(s, 0)))))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(merge, xs)
LIST2 → APP(s, app(s, app(s, 0)))
HAMMING → APP(s, 0)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(eq, x), y)
LIST2 → APP(s, app(s, 0))
LIST1 → APP(mult, app(s, app(s, 0)))
LIST2 → APP(mult, app(s, app(s, app(s, 0))))
HAMMING → APP(app(merge, list1), app(app(merge, list2), list3))
HAMMING → LIST2
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys)))
LIST1 → APP(app(map, app(mult, app(s, app(s, 0)))), hamming)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), ys)
APP(app(lt, app(s, x)), app(s, y)) → APP(lt, x)
HAMMING → APP(cons, app(s, 0))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(mult, app(s, x)), y) → APP(app(mult, x), y)
LIST2 → HAMMING
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(if, app(app(eq, x), y))
APP(app(plus, app(s, x)), y) → APP(plus, x)
LIST3 → APP(s, app(s, 0))
APP(app(lt, app(s, x)), app(s, y)) → APP(app(lt, x), y)
LIST1 → APP(s, app(s, 0))
LIST3 → APP(s, app(s, app(s, app(s, 0))))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(if, app(app(lt, x), y))
HAMMING → APP(app(merge, list2), list3)
HAMMING → LIST1
LIST2 → APP(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
LIST3 → APP(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0)))))))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
LIST3 → APP(mult, app(s, app(s, app(s, app(s, app(s, 0))))))
HAMMING → APP(merge, list1)
LIST3 → APP(s, app(s, app(s, 0)))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
LIST3 → APP(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
LIST2 → APP(map, app(mult, app(s, app(s, app(s, 0)))))
LIST1 → APP(s, 0)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(cons, x), app(app(merge, xs), ys))

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), app(app(cons, y), ys))
LIST2 → APP(s, 0)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys))))
LIST3 → HAMMING
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(lt, x)
HAMMING → APP(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))
HAMMING → LIST3
LIST3 → APP(s, 0)
LIST1 → HAMMING
APP(app(mult, app(s, x)), y) → APP(app(plus, y), app(app(mult, x), y))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys)))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(eq, x)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))
LIST1 → APP(map, app(mult, app(s, app(s, 0))))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(mult, app(s, x)), y) → APP(plus, y)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(lt, x), y)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, app(app(cons, x), xs)), ys)
HAMMING → APP(merge, list2)
APP(app(mult, app(s, x)), y) → APP(mult, x)
LIST3 → APP(s, app(s, app(s, app(s, app(s, 0)))))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(merge, xs)
LIST2 → APP(s, app(s, app(s, 0)))
HAMMING → APP(s, 0)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(eq, x), y)
LIST2 → APP(s, app(s, 0))
LIST1 → APP(mult, app(s, app(s, 0)))
LIST2 → APP(mult, app(s, app(s, app(s, 0))))
HAMMING → APP(app(merge, list1), app(app(merge, list2), list3))
HAMMING → LIST2
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys)))
LIST1 → APP(app(map, app(mult, app(s, app(s, 0)))), hamming)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), ys)
APP(app(lt, app(s, x)), app(s, y)) → APP(lt, x)
HAMMING → APP(cons, app(s, 0))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(mult, app(s, x)), y) → APP(app(mult, x), y)
LIST2 → HAMMING
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(if, app(app(eq, x), y))
APP(app(plus, app(s, x)), y) → APP(plus, x)
LIST3 → APP(s, app(s, 0))
APP(app(lt, app(s, x)), app(s, y)) → APP(app(lt, x), y)
LIST1 → APP(s, app(s, 0))
LIST3 → APP(s, app(s, app(s, app(s, 0))))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(if, app(app(lt, x), y))
HAMMING → APP(app(merge, list2), list3)
HAMMING → LIST1
LIST2 → APP(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
LIST3 → APP(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0)))))))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
LIST3 → APP(mult, app(s, app(s, app(s, app(s, app(s, 0))))))
HAMMING → APP(merge, list1)
LIST3 → APP(s, app(s, app(s, 0)))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
LIST3 → APP(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
LIST2 → APP(map, app(mult, app(s, app(s, app(s, 0)))))
LIST1 → APP(s, 0)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(cons, x), app(app(merge, xs), ys))

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 6 SCCs with 48 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)

R is empty.
The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

list1
list2
list3
hamming



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ ATransformationProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)

R is empty.
The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)

We have to consider all minimal (P,Q,R)-chains.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem. 

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ ATransformationProof
                          ↳ QDP
                            ↳ QReductionProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

plus1(s(x), y) → plus1(x, y)

R is empty.
The set Q consists of the following terms:

if(true, x0, x1)
if(false, x0, x1)
lt(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
eq(x0, x0)
eq(s(x0), 0)
eq(0, s(x0))
merge(x0, nil)
merge(nil, x0)
merge(cons(x0, x1), cons(x2, x3))
map(x0, nil)
map(x0, cons(x1, x2))
mult(0, x0)
mult(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

if(true, x0, x1)
if(false, x0, x1)
lt(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
eq(x0, x0)
eq(s(x0), 0)
eq(0, s(x0))
merge(x0, nil)
merge(nil, x0)
merge(cons(x0, x1), cons(x2, x3))
map(x0, nil)
map(x0, cons(x1, x2))
mult(0, x0)
mult(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ ATransformationProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ QDPSizeChangeProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

plus1(s(x), y) → plus1(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP(app(mult, app(s, x)), y) → APP(app(mult, x), y)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP(app(mult, app(s, x)), y) → APP(app(mult, x), y)

R is empty.
The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

list1
list2
list3
hamming



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ ATransformationProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP(app(mult, app(s, x)), y) → APP(app(mult, x), y)

R is empty.
The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)

We have to consider all minimal (P,Q,R)-chains.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem. 

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ ATransformationProof
                          ↳ QDP
                            ↳ QReductionProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

mult1(s(x), y) → mult1(x, y)

R is empty.
The set Q consists of the following terms:

if(true, x0, x1)
if(false, x0, x1)
lt(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
eq(x0, x0)
eq(s(x0), 0)
eq(0, s(x0))
merge(x0, nil)
merge(nil, x0)
merge(cons(x0, x1), cons(x2, x3))
map(x0, nil)
map(x0, cons(x1, x2))
mult(0, x0)
mult(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

if(true, x0, x1)
if(false, x0, x1)
lt(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
eq(x0, x0)
eq(s(x0), 0)
eq(0, s(x0))
merge(x0, nil)
merge(nil, x0)
merge(cons(x0, x1), cons(x2, x3))
map(x0, nil)
map(x0, cons(x1, x2))
mult(0, x0)
mult(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ ATransformationProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ QDPSizeChangeProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

mult1(s(x), y) → mult1(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP(app(lt, app(s, x)), app(s, y)) → APP(app(lt, x), y)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP(app(lt, app(s, x)), app(s, y)) → APP(app(lt, x), y)

R is empty.
The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

list1
list2
list3
hamming



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ ATransformationProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP(app(lt, app(s, x)), app(s, y)) → APP(app(lt, x), y)

R is empty.
The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)

We have to consider all minimal (P,Q,R)-chains.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem. 

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ ATransformationProof
                          ↳ QDP
                            ↳ QReductionProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

lt1(s(x), s(y)) → lt1(x, y)

R is empty.
The set Q consists of the following terms:

if(true, x0, x1)
if(false, x0, x1)
lt(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
eq(x0, x0)
eq(s(x0), 0)
eq(0, s(x0))
merge(x0, nil)
merge(nil, x0)
merge(cons(x0, x1), cons(x2, x3))
map(x0, nil)
map(x0, cons(x1, x2))
mult(0, x0)
mult(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

if(true, x0, x1)
if(false, x0, x1)
lt(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
eq(x0, x0)
eq(s(x0), 0)
eq(0, s(x0))
merge(x0, nil)
merge(nil, x0)
merge(cons(x0, x1), cons(x2, x3))
map(x0, nil)
map(x0, cons(x1, x2))
mult(0, x0)
mult(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ ATransformationProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ QDPSizeChangeProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

lt1(s(x), s(y)) → lt1(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), ys)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), app(app(cons, y), ys))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, app(app(cons, x), xs)), ys)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), ys)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), app(app(cons, y), ys))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, app(app(cons, x), xs)), ys)

R is empty.
The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

list1
list2
list3
hamming



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ ATransformationProof
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), ys)
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, xs), app(app(cons, y), ys))
APP(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → APP(app(merge, app(app(cons, x), xs)), ys)

R is empty.
The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)

We have to consider all minimal (P,Q,R)-chains.
We have applied the A-Transformation [17] to get from an applicative problem to a standard problem. 

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ ATransformationProof
                          ↳ QDP
                            ↳ QReductionProof
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

merge1(cons(x, xs), cons(y, ys)) → merge1(xs, cons(y, ys))
merge1(cons(x, xs), cons(y, ys)) → merge1(xs, ys)
merge1(cons(x, xs), cons(y, ys)) → merge1(cons(x, xs), ys)

R is empty.
The set Q consists of the following terms:

if(true, x0, x1)
if(false, x0, x1)
lt(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
eq(x0, x0)
eq(s(x0), 0)
eq(0, s(x0))
merge(x0, nil)
merge(nil, x0)
merge(cons(x0, x1), cons(x2, x3))
map(x0, nil)
map(x0, cons(x1, x2))
mult(0, x0)
mult(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

if(true, x0, x1)
if(false, x0, x1)
lt(s(x0), s(x1))
lt(0, s(x0))
lt(x0, 0)
eq(x0, x0)
eq(s(x0), 0)
eq(0, s(x0))
merge(x0, nil)
merge(nil, x0)
merge(cons(x0, x1), cons(x2, x3))
map(x0, nil)
map(x0, cons(x1, x2))
mult(0, x0)
mult(s(x0), x1)
plus(0, x0)
plus(s(x0), x1)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ ATransformationProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ QDPSizeChangeProof
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

merge1(cons(x, xs), cons(y, ys)) → merge1(xs, cons(y, ys))
merge1(cons(x, xs), cons(y, ys)) → merge1(xs, ys)
merge1(cons(x, xs), cons(y, ys)) → merge1(cons(x, xs), ys)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

R is empty.
The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

list1
list2
list3
hamming



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

R is empty.
The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)

We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof

Q DP problem:
The TRS P consists of the following rules:

HAMMING → LIST3
LIST2 → HAMMING
HAMMING → LIST1
HAMMING → LIST2
LIST1 → HAMMING
LIST3 → HAMMING

The TRS R consists of the following rules:

app(app(app(if, true), xs), ys) → xs
app(app(app(if, false), xs), ys) → ys
app(app(lt, app(s, x)), app(s, y)) → app(app(lt, x), y)
app(app(lt, 0), app(s, y)) → true
app(app(lt, y), 0) → false
app(app(eq, x), x) → true
app(app(eq, app(s, x)), 0) → false
app(app(eq, 0), app(s, x)) → false
app(app(merge, xs), nil) → xs
app(app(merge, nil), ys) → ys
app(app(merge, app(app(cons, x), xs)), app(app(cons, y), ys)) → app(app(app(if, app(app(lt, x), y)), app(app(cons, x), app(app(merge, xs), app(app(cons, y), ys)))), app(app(app(if, app(app(eq, x), y)), app(app(cons, x), app(app(merge, xs), ys))), app(app(cons, y), app(app(merge, app(app(cons, x), xs)), ys))))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(mult, 0), x) → 0
app(app(mult, app(s, x)), y) → app(app(plus, y), app(app(mult, x), y))
app(app(plus, 0), x) → 0
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
list1 → app(app(map, app(mult, app(s, app(s, 0)))), hamming)
list2 → app(app(map, app(mult, app(s, app(s, app(s, 0))))), hamming)
list3 → app(app(map, app(mult, app(s, app(s, app(s, app(s, app(s, 0))))))), hamming)
hamming → app(app(cons, app(s, 0)), app(app(merge, list1), app(app(merge, list2), list3)))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof

Q DP problem:
The TRS P consists of the following rules:

HAMMING → LIST3
LIST2 → HAMMING
HAMMING → LIST1
HAMMING → LIST2
LIST1 → HAMMING
LIST3 → HAMMING

R is empty.
The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

app(app(app(if, true), x0), x1)
app(app(app(if, false), x0), x1)
app(app(lt, app(s, x0)), app(s, x1))
app(app(lt, 0), app(s, x0))
app(app(lt, x0), 0)
app(app(eq, x0), x0)
app(app(eq, app(s, x0)), 0)
app(app(eq, 0), app(s, x0))
app(app(merge, x0), nil)
app(app(merge, nil), x0)
app(app(merge, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(mult, 0), x0)
app(app(mult, app(s, x0)), x1)
app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
list1
list2
list3
hamming



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ NonTerminationProof

Q DP problem:
The TRS P consists of the following rules:

LIST2 → HAMMING
HAMMING → LIST3
HAMMING → LIST1
HAMMING → LIST2
LIST1 → HAMMING
LIST3 → HAMMING

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

LIST2 → HAMMING
HAMMING → LIST3
HAMMING → LIST1
HAMMING → LIST2
LIST1 → HAMMING
LIST3 → HAMMING

The TRS R consists of the following rules:none


s = HAMMING evaluates to t =HAMMING

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

HAMMING → LIST2
with rule HAMMING → LIST2 at position [] and matcher [ ]

LIST2 → HAMMING
with rule LIST2 → HAMMING

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.