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) app(pred,app(s,x)) -> x
(2) app(app(minus,x),0) -> x
(3) app(app(minus,x),app(s,y)) -> app(pred,app(app(minus,x),y))
(4) app(app(quot,0),app(s,y)) -> 0
(5) app(app(quot,app(s,x)),app(s,y)) -> app(s,app(app(quot,app(app(minus,x),y)),app(s,y)))

[2]  Apply uncurrying transformation with <app> as function application symbol to obtain the following TRS:
(1) pred(s(x)) -> x
(2) minus(x,0) -> x
(3) minus(x,s(y)) -> pred(minus(x,y))
(4) quot(0,s(y)) -> 0
(5) quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))

[3]  Apply the Dependency Pair transformation resulting in the following dependency pairs:
<Minus(x,s(y)), Pred(minus(x,y))>
<Minus(x,s(y)), Minus(x,y)>
<Quot(s(x),s(y)), Quot(minus(x,y),s(y))>
<Quot(s(x),s(y)), Minus(x,y)>
 and 2 SCCs in the dependency graph. Proofs for those SCCs follow.

[3a-1]  Consider the following SCC obtained from analysis of dependency graph:
(1) pred(s(x)) ->= x
(2) minus(x,0) ->= x
(3) minus(x,s(y)) ->= pred(minus(x,y))
(4) quot(0,s(y)) ->= 0
(5) quot(s(x),s(y)) ->= s(quot(minus(x,y),s(y)))
(DP3) Quot(s(x),s(y)) -> Quot(minus(x,y),s(y))

[3a-2]  Use following polynomial interpretation:
[s(x)] = x + 1
[minus(x,y)] = x
rest default

  Remove rules with left hand side strictly bigger than right hand side:
(1), (4), (DP3)

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

[3b-1]  Consider the following SCC obtained from analysis of dependency graph:
(1) pred(s(x)) ->= x
(2) minus(x,0) ->= x
(3) minus(x,s(y)) ->= pred(minus(x,y))
(4) quot(0,s(y)) ->= 0
(5) quot(s(x),s(y)) ->= s(quot(minus(x,y),s(y)))
(DP2) Minus(x,s(y)) -> Minus(x,y)

[3b-2]  Use following polynomial interpretation:
[s(x)] = x + 1
[minus(x,y)] = x
rest default

  Remove rules with left hand side strictly bigger than right hand side:
(1), (4), (DP2)

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



../tpdb/TRS/currying/AG01/#3.2.trs, 0.01, Y
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