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) plus(0,Y) -> Y
(2) plus(s(X),Y) -> s(plus(X,Y))
(3) min(X,0) -> X
(4) min(s(X),s(Y)) -> min(X,Y)
(5) min(min(X,Y),Z) -> min(X,plus(Y,Z))
(6) quot(0,s(Y)) -> 0
(7) quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y)))

[2]  Apply the Dependency Pair transformation resulting in the following dependency pairs:
<Plus(s(X),Y), Plus(X,Y)>
<Min(s(X),s(Y)), Min(X,Y)>
<Min(min(X,Y),Z), Min(X,plus(Y,Z))>
<Min(min(X,Y),Z), Plus(Y,Z)>
<Quot(s(X),s(Y)), Quot(min(X,Y),s(Y))>
<Quot(s(X),s(Y)), Min(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) plus(0,Y) ->= Y
(2) plus(s(X),Y) ->= s(plus(X,Y))
(3) min(X,0) ->= X
(4) min(s(X),s(Y)) ->= min(X,Y)
(5) min(min(X,Y),Z) ->= min(X,plus(Y,Z))
(6) quot(0,s(Y)) ->= 0
(7) quot(s(X),s(Y)) ->= s(quot(min(X,Y),s(Y)))
(DP5) Quot(s(X),s(Y)) -> Quot(min(X,Y),s(Y))

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

  Remove rules with left hand side strictly bigger than right hand side:
(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) plus(0,Y) ->= Y
(2) plus(s(X),Y) ->= s(plus(X,Y))
(3) min(X,0) ->= X
(4) min(s(X),s(Y)) ->= min(X,Y)
(5) min(min(X,Y),Z) ->= min(X,plus(Y,Z))
(6) quot(0,s(Y)) ->= 0
(7) quot(s(X),s(Y)) ->= s(quot(min(X,Y),s(Y)))
(DP3) Min(min(X,Y),Z) -> Min(X,plus(Y,Z))
(DP2) Min(s(X),s(Y)) -> Min(X,Y)

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

  Remove rules with left hand side strictly bigger than right hand side:
(3)-(5), (DP3), (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) plus(0,Y) ->= Y
(2) plus(s(X),Y) ->= s(plus(X,Y))
(3) min(X,0) ->= X
(4) min(s(X),s(Y)) ->= min(X,Y)
(5) min(min(X,Y),Z) ->= min(X,plus(Y,Z))
(6) quot(0,s(Y)) ->= 0
(7) quot(s(X),s(Y)) ->= s(quot(min(X,Y),s(Y)))
(DP1) Plus(s(X),Y) -> Plus(X,Y)

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

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

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



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