(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) RelTRStoRelADPProof (EQUIVALENT transformation)
We upgrade the RelTRS problem to an equivalent Relative ADP Problem [IJCAR24].
(2) Obligation:
Relative ADP Problem with
absolute ADPs:
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))
and relative ADPs:
rand(x) → RAND(s(x))
rand(x) → x
(3) RelADPDepGraphProof (EQUIVALENT transformation)
We use the relative dependency graph processor [IJCAR24].
The approximation of the Relative Dependency Graph contains:
1 SCC with nodes from P_abs,
0 Lassos,
Result: This relative DT problem is equivalent to 1 subproblem.
(4) Obligation:
Relative ADP Problem with
absolute ADPs:
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))
and relative ADPs:
rand(x) → rand(s(x))
rand(x) → x
(5) RelADPDerelatifyingProof (EQUIVALENT transformation)
We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(c(x, s(y))) → G(c(s(x), y))
F(c(s(x), s(y))) → G(c(x, y))
F(c(s(x), y)) → F(c(x, s(y)))
G(c(s(x), s(y))) → F(c(x, y))
The TRS R consists of the following rules:
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))
rand(x) → rand(s(x))
f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
rand(x) → x
Q is empty.
We have to consider all (P,Q,R)-chains.
(7) 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(F(x1)) = 2 + 2·x1
POL(G(x1)) = 2 + 2·x1
POL(c(x1, x2)) = 2 + 2·x1 + x2
POL(f(x1)) = 2·x1
POL(g(x1)) = 2·x1
POL(rand(x1)) = 2 + x1
POL(s(x1)) = x1
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(c(x, s(y))) → G(c(s(x), y))
F(c(s(x), s(y))) → G(c(x, y))
F(c(s(x), y)) → F(c(x, s(y)))
G(c(s(x), s(y))) → F(c(x, y))
The TRS R consists of the following rules:
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))
rand(x) → rand(s(x))
f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
Q is empty.
We have to consider all (P,Q,R)-chains.
(9) 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( g(x1) ) = max{0, -2} |
| POL( c(x1, x2) ) = 2x1 + 2x2 + 1 |
| POL( rand(x1) ) = max{0, -2} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))
rand(x) → rand(s(x))
f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(c(x, s(y))) → G(c(s(x), y))
F(c(s(x), y)) → F(c(x, s(y)))
The TRS R consists of the following rules:
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))
rand(x) → rand(s(x))
f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
Q is empty.
We have to consider all (P,Q,R)-chains.
(11) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
(12) Complex Obligation (AND)
(13) 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:
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))
rand(x) → rand(s(x))
f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
Q is empty.
We have to consider all (P,Q,R)-chains.
(14) 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: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( c(x1, x2) ) = 2x1 + x2 + 1 |
| POL( rand(x1) ) = max{0, -2} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))
rand(x) → rand(s(x))
f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
(15) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))
rand(x) → rand(s(x))
f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
Q is empty.
We have to consider all (P,Q,R)-chains.
(16) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(17) YES
(18) 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:
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))
rand(x) → rand(s(x))
f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
Q is empty.
We have to consider all (P,Q,R)-chains.
(19) 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: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( g(x1) ) = max{0, -1} |
| POL( c(x1, x2) ) = 2x2 + 2 |
| POL( rand(x1) ) = max{0, -2} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))
rand(x) → rand(s(x))
f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
(20) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))
rand(x) → rand(s(x))
f(c(s(x), y)) → f(c(x, s(y)))
f(c(s(x), s(y))) → g(c(x, y))
Q is empty.
We have to consider all (P,Q,R)-chains.
(21) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(22) YES