YES Termination w.r.t. Q proof of ../tpdb/TRS/Zantema06/beans.trs
Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.

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

L1(f(s(s(y)), f(z, w))) → F(s(z), w)
L1(f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), nil)) → F(s(0), nil)
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
L1(f(s(s(y)), f(z, w))) → F(s(0), f(y, f(s(z), w)))
F(x, f(s(s(y)), nil)) → F(y, f(s(0), nil))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

L1(f(s(s(y)), f(z, w))) → F(s(z), w)
L1(f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), nil)) → F(s(0), nil)
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
L1(f(s(s(y)), f(z, w))) → F(s(0), f(y, f(s(z), w)))
F(x, f(s(s(y)), nil)) → F(y, f(s(0), nil))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
          ↳ QDP

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

F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ RuleRemovalProof
          ↳ QDP

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

F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(F(x1, x2)) = x1 + 2·x2   
POL(f(x1, x2)) = 2·x1 + x2   
POL(nil) = 0   
POL(s(x1)) = 1 + x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ RuleRemovalProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof

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

L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ Narrowing

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

L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w)))) at position [0] we obtained the following new rules:

L1(f(s(s(y0)), f(y1, f(s(s(x1)), f(x2, x3))))) → L1(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(x2), x3))))))
L1(f(s(s(x0)), f(s(x1), f(x2, x3)))) → L1(f(s(0), f(s(x0), f(x1, f(s(x2), x3)))))
L1(f(s(s(x0)), f(s(x1), nil))) → L1(f(s(0), f(s(x0), f(x1, f(s(0), nil)))))
L1(f(s(s(y0)), f(y1, f(s(s(x1)), nil)))) → L1(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(0), nil))))))
L1(f(s(s(s(s(x1)))), f(y1, x3))) → L1(f(s(s(0)), f(x1, f(s(s(y1)), x3))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof

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

L1(f(s(s(y0)), f(y1, f(s(s(x1)), f(x2, x3))))) → L1(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(x2), x3))))))
L1(f(s(s(x0)), f(s(x1), nil))) → L1(f(s(0), f(s(x0), f(x1, f(s(0), nil)))))
L1(f(s(s(x0)), f(s(x1), f(x2, x3)))) → L1(f(s(0), f(s(x0), f(x1, f(s(x2), x3)))))
L1(f(s(s(s(s(x1)))), f(y1, x3))) → L1(f(s(s(0)), f(x1, f(s(s(y1)), x3))))
L1(f(s(s(y0)), f(y1, f(s(s(x1)), nil)))) → L1(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(0), nil))))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ QDP
                        ↳ QDPOrderProof

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

L1(f(s(s(y0)), f(y1, f(s(s(x1)), f(x2, x3))))) → L1(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(x2), x3))))))
L1(f(s(s(x0)), f(s(x1), f(x2, x3)))) → L1(f(s(0), f(s(x0), f(x1, f(s(x2), x3)))))
L1(f(s(s(s(s(x1)))), f(y1, x3))) → L1(f(s(s(0)), f(x1, f(s(s(y1)), x3))))
L1(f(s(s(y0)), f(y1, f(s(s(x1)), nil)))) → L1(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(0), nil))))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


L1(f(s(s(y0)), f(y1, f(s(s(x1)), f(x2, x3))))) → L1(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(x2), x3))))))
L1(f(s(s(x0)), f(s(x1), f(x2, x3)))) → L1(f(s(0), f(s(x0), f(x1, f(s(x2), x3)))))
L1(f(s(s(s(s(x1)))), f(y1, x3))) → L1(f(s(s(0)), f(x1, f(s(s(y1)), x3))))
L1(f(s(s(y0)), f(y1, f(s(s(x1)), nil)))) → L1(f(s(0), f(y0, f(s(s(y1)), f(x1, f(s(0), nil))))))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [21,25] with Interpretation:

POL( s(x1) ) = x1 + 2


POL( 0 ) = max{0, -1}


POL( nil ) = 1


POL( f(x1, x2) ) = max{0, x1 + x2 - 1}


POL( L1(x1) ) = max{0, x1 - 1}



The following usable rules [17] were oriented:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.