(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))
Q is empty.
(1) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))
The set Q consists of the following terms:
ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))
(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:
AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(ap(g, x), y), ap(ap(x, y), 0))
AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(x, y), 0)
AP(ap(ap(g, x), y), ap(s, z)) → AP(x, y)
The TRS R consists of the following rules:
ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))
The set Q consists of the following terms:
ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
AP(ap(ap(g, x), y), ap(s, z)) → AP(x, y)
AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(ap(g, x), y), ap(ap(x, y), 0))
The TRS R consists of the following rules:
ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))
The set Q consists of the following terms:
ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))
We have to consider all minimal (P,Q,R)-chains.
(7) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
AP(ap(ap(g, x), y), ap(s, z)) → AP(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
AP(
x1,
x2) =
x1
ap(
x1,
x2) =
ap(
x1,
x2)
g =
g
Recursive Path Order [RPO].
Precedence:
trivial
The following usable rules [FROCOS05] were oriented:
none
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
AP(ap(ap(g, x), y), ap(s, z)) → AP(ap(ap(g, x), y), ap(ap(x, y), 0))
The TRS R consists of the following rules:
ap(f, x) → x
ap(ap(ap(g, x), y), ap(s, z)) → ap(ap(ap(g, x), y), ap(ap(x, y), 0))
The set Q consists of the following terms:
ap(f, x0)
ap(ap(ap(g, x0), x1), ap(s, x2))
We have to consider all minimal (P,Q,R)-chains.