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

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

[3]  Apply the Dependency Pair transformation resulting in the following dependency pairs:
<Minus(s(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)>
<Log(s(s(x))), Log(s(quot(x,s(s(0)))))>
<Log(s(s(x))), Quot(x,s(s(0)))>
 and 3 SCCs in the dependency graph. Proofs for those SCCs follow.

[3a-1]  Consider the following SCC obtained from analysis of dependency graph:
(1) minus(x,0) ->= x
(2) minus(s(x),s(y)) ->= minus(x,y)
(3) quot(0,s(y)) ->= 0
(4) quot(s(x),s(y)) ->= s(quot(minus(x,y),s(y)))
(5) log(s(0)) ->= 0
(6) log(s(s(x))) ->= s(log(s(quot(x,s(s(0))))))
(DP4) Log(s(s(x))) -> Log(s(quot(x,s(s(0)))))

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

  Remove rules with left hand side strictly bigger than right hand side:
(2), (5), (DP4)

[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) minus(x,0) ->= x
(2) minus(s(x),s(y)) ->= minus(x,y)
(3) quot(0,s(y)) ->= 0
(4) quot(s(x),s(y)) ->= s(quot(minus(x,y),s(y)))
(5) log(s(0)) ->= 0
(6) log(s(s(x))) ->= s(log(s(quot(x,s(s(0))))))
(DP2) Quot(s(x),s(y)) -> Quot(minus(x,y),s(y))

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

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

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

[3c-1]  Consider the following SCC obtained from analysis of dependency graph:
(1) minus(x,0) ->= x
(2) minus(s(x),s(y)) ->= minus(x,y)
(3) quot(0,s(y)) ->= 0
(4) quot(s(x),s(y)) ->= s(quot(minus(x,y),s(y)))
(5) log(s(0)) ->= 0
(6) log(s(s(x))) ->= s(log(s(quot(x,s(s(0))))))
(DP1) Minus(s(x),s(y)) -> Minus(x,y)

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

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

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



../tpdb/TRS/currying/AG01/#3.8.trs, 0.01, Y
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