(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
Q is empty.
(1) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The TRS R 2 is
f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
The signature Sigma is {
f}
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
f(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
F(true, x, y) → GT(x, y)
GT(s(u), s(v)) → GT(u, v)
The TRS R consists of the following rules:
f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
f(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GT(s(u), s(v)) → GT(u, v)
The TRS R consists of the following rules:
f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
f(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(8) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
GT(s(u), s(v)) → GT(u, v)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
GT(
x1,
x2) =
x2
s(
x1) =
s(
x1)
f(
x1,
x2,
x3) =
f
true =
true
gt(
x1,
x2) =
gt
0 =
0
false =
false
Recursive path order with status [RPO].
Precedence:
gt > false
gt > true
Status:
trivial
The following usable rules [FROCOS05] were oriented:
f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
(9) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
f(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(10) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(11) TRUE
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
The TRS R consists of the following rules:
f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
The set Q consists of the following terms:
f(true, x0, x1)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.