YES
  TPA v.1.0

Result: TRS is terminating

Default interpretations for symbols are not printed. For polynomial interpretations
and semantic labelling over N\{0,1} defaults are 2 for constants, identity for unary
symbols and x+y-2 for binary symbols. For semantic labelling over {0,1} (booleans)
defaults are 0 for constants, identity for unary symbols and disjunction for binary symbols.

[1]  TRS loaded from input file:
(1) -(x,0) -> x
(2) -(0,s(y)) -> 0
(3) -(s(x),s(y)) -> -(x,y)
(4) lt(x,0) -> false
(5) lt(0,s(y)) -> true
(6) lt(s(x),s(y)) -> lt(x,y)
(7) if`1(true,if`2(x,y)) -> x
(8) if`1(false,if`2(x,y)) -> y
(9) div(x,0) -> 0
(10) div(0,y) -> 0
(11) div(s(x),s(y)) -> if`1(lt(x,y),if`2(0,s(div(-(x,y),s(y)))))

[2]  Apply the Dependency Pair transformation resulting in the following dependency pairs:
<-~(s(x),s(y)), -~(x,y)>
<Lt(s(x),s(y)), Lt(x,y)>
<Div(s(x),s(y)), If`1(lt(x,y),if`2(0,s(div(-(x,y),s(y)))))>
<Div(s(x),s(y)), Lt(x,y)>
<Div(s(x),s(y)), Div(-(x,y),s(y))>
<Div(s(x),s(y)), -~(x,y)>
 and 3 SCCs in the dependency graph. Proofs for those SCCs follow.

[2a-1]  Consider the following SCC obtained from analysis of dependency graph:
(1) -(x,0) ->= x
(2) -(0,s(y)) ->= 0
(3) -(s(x),s(y)) ->= -(x,y)
(4) lt(x,0) ->= false
(5) lt(0,s(y)) ->= true
(6) lt(s(x),s(y)) ->= lt(x,y)
(7) if`1(true,if`2(x,y)) ->= x
(8) if`1(false,if`2(x,y)) ->= y
(9) div(x,0) ->= 0
(10) div(0,y) ->= 0
(11) div(s(x),s(y)) ->= if`1(lt(x,y),if`2(0,s(div(-(x,y),s(y)))))
(DP5) Div(s(x),s(y)) -> Div(-(x,y),s(y))

[2a-2]  Use following polynomial interpretation:
[-(x,y)] = x
[s(x)] = x + 1
[if`1(x,y)] = y
rest default

  Remove rules with left hand side strictly bigger than right hand side:
(3), (5)-(6), (DP5)

[2a-3]  Since there are no remaining strict rules, relative termination is proved!

[2b-1]  Consider the following SCC obtained from analysis of dependency graph:
(1) -(x,0) ->= x
(2) -(0,s(y)) ->= 0
(3) -(s(x),s(y)) ->= -(x,y)
(4) lt(x,0) ->= false
(5) lt(0,s(y)) ->= true
(6) lt(s(x),s(y)) ->= lt(x,y)
(7) if`1(true,if`2(x,y)) ->= x
(8) if`1(false,if`2(x,y)) ->= y
(9) div(x,0) ->= 0
(10) div(0,y) ->= 0
(11) div(s(x),s(y)) ->= if`1(lt(x,y),if`2(0,s(div(-(x,y),s(y)))))
(DP2) Lt(s(x),s(y)) -> Lt(x,y)

[2b-2]  Use following polynomial interpretation:
[-(x,y)] = x
[s(x)] = x + 1
[if`1(x,y)] = y
rest default

  Remove rules with left hand side strictly bigger than right hand side:
(3), (5)-(6), (DP2)

[2b-3]  Since there are no remaining strict rules, relative termination is proved!

[2c-1]  Consider the following SCC obtained from analysis of dependency graph:
(1) -(x,0) ->= x
(2) -(0,s(y)) ->= 0
(3) -(s(x),s(y)) ->= -(x,y)
(4) lt(x,0) ->= false
(5) lt(0,s(y)) ->= true
(6) lt(s(x),s(y)) ->= lt(x,y)
(7) if`1(true,if`2(x,y)) ->= x
(8) if`1(false,if`2(x,y)) ->= y
(9) div(x,0) ->= 0
(10) div(0,y) ->= 0
(11) div(s(x),s(y)) ->= if`1(lt(x,y),if`2(0,s(div(-(x,y),s(y)))))
(DP1) -~(s(x),s(y)) -> -~(x,y)

[2c-2]  Use following polynomial interpretation:
[-(x,y)] = x
[s(x)] = x + 1
[if`1(x,y)] = y
rest default

  Remove rules with left hand side strictly bigger than right hand side:
(3), (5)-(6), (DP1)

[2c-3]  Since there are no remaining strict rules, relative termination is proved!



../tpdb/TRS/HM/t014.trs, 0.02, Y
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