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) minus(0,Y) -> 0
(2) minus(s(X),s(Y)) -> minus(X,Y)
(3) geq(X,0) -> true
(4) geq(0,s(Y)) -> false
(5) geq(s(X),s(Y)) -> geq(X,Y)
(6) div(0,s(Y)) -> 0
(7) div(s(X),s(Y)) -> if`1(geq(X,Y),if`2(s(div(minus(X,Y),s(Y))),0))
(8) if`1(true,if`2(X,Y)) -> X
(9) if`1(false,if`2(X,Y)) -> Y

[2]  Apply the Dependency Pair transformation resulting in the following dependency pairs:
<Minus(s(X),s(Y)), Minus(X,Y)>
<Geq(s(X),s(Y)), Geq(X,Y)>
<Div(s(X),s(Y)), If`1(geq(X,Y),if`2(s(div(minus(X,Y),s(Y))),0))>
<Div(s(X),s(Y)), Geq(X,Y)>
<Div(s(X),s(Y)), Div(minus(X,Y),s(Y))>
<Div(s(X),s(Y)), Minus(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) minus(0,Y) ->= 0
(2) minus(s(X),s(Y)) ->= minus(X,Y)
(3) geq(X,0) ->= true
(4) geq(0,s(Y)) ->= false
(5) geq(s(X),s(Y)) ->= geq(X,Y)
(6) div(0,s(Y)) ->= 0
(7) div(s(X),s(Y)) ->= if`1(geq(X,Y),if`2(s(div(minus(X,Y),s(Y))),0))
(8) if`1(true,if`2(X,Y)) ->= X
(9) if`1(false,if`2(X,Y)) ->= Y
(DP5) Div(s(X),s(Y)) -> Div(minus(X,Y),s(Y))

[2a-2]  Use following polynomial interpretation:
[minus(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:
(2), (4)-(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) minus(0,Y) ->= 0
(2) minus(s(X),s(Y)) ->= minus(X,Y)
(3) geq(X,0) ->= true
(4) geq(0,s(Y)) ->= false
(5) geq(s(X),s(Y)) ->= geq(X,Y)
(6) div(0,s(Y)) ->= 0
(7) div(s(X),s(Y)) ->= if`1(geq(X,Y),if`2(s(div(minus(X,Y),s(Y))),0))
(8) if`1(true,if`2(X,Y)) ->= X
(9) if`1(false,if`2(X,Y)) ->= Y
(DP2) Geq(s(X),s(Y)) -> Geq(X,Y)

[2b-2]  Use following polynomial interpretation:
[minus(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:
(2), (4)-(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) minus(0,Y) ->= 0
(2) minus(s(X),s(Y)) ->= minus(X,Y)
(3) geq(X,0) ->= true
(4) geq(0,s(Y)) ->= false
(5) geq(s(X),s(Y)) ->= geq(X,Y)
(6) div(0,s(Y)) ->= 0
(7) div(s(X),s(Y)) ->= if`1(geq(X,Y),if`2(s(div(minus(X,Y),s(Y))),0))
(8) if`1(true,if`2(X,Y)) ->= X
(9) if`1(false,if`2(X,Y)) ->= Y
(DP1) Minus(s(X),s(Y)) -> Minus(X,Y)

[2c-2]  Use following polynomial interpretation:
[minus(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:
(2), (4)-(6), (DP1)

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



../tpdb/TRS/nontermin/CSR/Ex49_GM04.trs, 0.01, Y
Couldn't open file <60>: 60: No such file or directory