(0) Obligation:
Relative term rewrite system:
The relative TRS consists of the following R rules:
f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))
The relative TRS consists of the following S rules:
rand(x) → rand(s(x))
rand(x) → x
(1) RelTRStoQDPProof (SOUND transformation)
The relative termination problem is root-restricted. We can therefore treat it as a dependency pair problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))
The TRS R consists of the following rules:
rand(x) → rand(s(x))
rand(x) → x
Q is empty.
We have to consider all (P,Q,R)-chains.
(3) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented rules of the TRS R:
rand(x) → x
Used ordering: Polynomial interpretation [POLO]:
POL(c(x1, x2)) = 2·x1 + x2
POL(f(x1)) = 2·x1
POL(g(x1)) = 2·x1
POL(rand(x1)) = 2 + x1
POL(s(x1)) = x1
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))
The TRS R consists of the following rules:
rand(x) → rand(s(x))
Q is empty.
We have to consider all (P,Q,R)-chains.
(5) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
f(c(s(x), s(y))) → g(c(x, y))
g(c(s(x), s(y))) → f(c(x, y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( c(x1, x2) ) = 2x1 + 2x2 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
rand(x) → rand(s(x))
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))
The TRS R consists of the following rules:
rand(x) → rand(s(x))
Q is empty.
We have to consider all (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
(8) Complex Obligation (AND)
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
g(c(x, s(y))) → g(c(s(x), y))
The TRS R consists of the following rules:
rand(x) → rand(s(x))
Q is empty.
We have to consider all (P,Q,R)-chains.
(10) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
g(c(x, s(y))) → g(c(s(x), y))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
g(
x1) =
x1
c(
x1,
x2) =
x2
s(
x1) =
s(
x1)
rand(
x1) =
rand
Knuth-Bendix order [KBO] with precedence:
trivial
and weight map:
s_1=1
rand=1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
rand(x) → rand(s(x))
(11) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
rand(x) → rand(s(x))
Q is empty.
We have to consider all (P,Q,R)-chains.
(12) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(13) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
f(c(s(x), y)) → f(c(x, s(y)))
The TRS R consists of the following rules:
rand(x) → rand(s(x))
Q is empty.
We have to consider all (P,Q,R)-chains.
(15) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
f(c(s(x), y)) → f(c(x, s(y)))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
f(
x1) =
x1
c(
x1,
x2) =
x1
s(
x1) =
s(
x1)
rand(
x1) =
rand
Knuth-Bendix order [KBO] with precedence:
trivial
and weight map:
s_1=1
rand=1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
rand(x) → rand(s(x))
(16) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
rand(x) → rand(s(x))
Q is empty.
We have to consider all (P,Q,R)-chains.
(17) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(18) YES