(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))
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:
f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))
The set Q consists of the following terms:
f(0, 1, x0)
f(x0, x1, 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:
F(0, 1, x) → F(s(x), x, x)
F(x, y, s(z)) → F(0, 1, z)
The TRS R consists of the following rules:
f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))
The set Q consists of the following terms:
f(0, 1, x0)
f(x0, x1, s(x2))
We have to consider all minimal (P,Q,R)-chains.
(5) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
F(x, y, s(z)) → F(0, 1, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(
x1,
x2,
x3) =
x3
0 =
0
1 =
1
s(
x1) =
s(
x1)
f(
x1,
x2,
x3) =
f(
x3)
Lexicographic path order with status [LPO].
Quasi-Precedence:
f1 > 0 > [1, s1]
Status:
f1: [1]
s1: [1]
1: []
0: []
The following usable rules [FROCOS05] were oriented:
f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, x) → F(s(x), x, x)
The TRS R consists of the following rules:
f(0, 1, x) → f(s(x), x, x)
f(x, y, s(z)) → s(f(0, 1, z))
The set Q consists of the following terms:
f(0, 1, x0)
f(x0, x1, s(x2))
We have to consider all minimal (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(8) TRUE