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(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) plus(0,y) -> y
(6) plus(s(x),y) -> s(plus(x,y))
(7) minus(minus(x,y),z) -> minus(x,plus(y,z))

[2]  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)>
<Plus(s(x),y), Plus(x,y)>
<Minus(minus(x,y),z), Minus(x,plus(y,z))>
<Minus(minus(x,y),z), Plus(y,z)>
 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(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) plus(0,y) ->= y
(6) plus(s(x),y) ->= s(plus(x,y))
(7) minus(minus(x,y),z) ->= minus(x,plus(y,z))
(DP2) Quot(s(x),s(y)) -> Quot(minus(x,y),s(y))

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

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

[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(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) plus(0,y) ->= y
(6) plus(s(x),y) ->= s(plus(x,y))
(7) minus(minus(x,y),z) ->= minus(x,plus(y,z))
(DP5) Minus(minus(x,y),z) -> Minus(x,plus(y,z))
(DP1) Minus(s(x),s(y)) -> Minus(x,y)

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

  Remove rules with left hand side strictly bigger than right hand side:
(1)-(2), (7), (DP5), (DP1)

[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(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) plus(0,y) ->= y
(6) plus(s(x),y) ->= s(plus(x,y))
(7) minus(minus(x,y),z) ->= minus(x,plus(y,z))
(DP4) Plus(s(x),y) -> Plus(x,y)

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

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

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



../tpdb/TRS/AG01/#3.4.trs, 0.01, Y
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