(0) Obligation:
Q restricted rewrite system:
The TRS R 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))
g(s(f(x))) → g(f(x))
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(c(x1, x2)) = x1 + x2
POL(f(x1)) = x1
POL(g(x1)) = x1
POL(s(x1)) = 1 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
g(s(f(x))) → g(f(x))
(2) Obligation:
Q restricted rewrite system:
The TRS R 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))
Q is empty.
(3) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))
The signature Sigma is {
f,
g}
(4) Obligation:
Q restricted rewrite system:
The TRS R 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 set Q consists of the following terms:
f(c(s(x0), x1))
g(c(x0, s(x1)))
(5) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(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:
f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))
The set Q consists of the following terms:
f(c(s(x0), x1))
g(c(x0, s(x1)))
We have to consider all minimal (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:
f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))
The set Q consists of the following terms:
f(c(s(x0), x1))
g(c(x0, s(x1)))
We have to consider all minimal (P,Q,R)-chains.
(10) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(c(x, s(y))) → G(c(s(x), y))
R is empty.
The set Q consists of the following terms:
f(c(s(x0), x1))
g(c(x0, s(x1)))
We have to consider all minimal (P,Q,R)-chains.
(12) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
f(c(s(x0), x1))
g(c(x0, s(x1)))
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(c(x, s(y))) → G(c(s(x), y))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(14) 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 dependency pairs:
G(c(x, s(y))) → G(c(s(x), y))
Used ordering: Polynomial interpretation [POLO]:
POL(G(x1)) = 2·x1
POL(c(x1, x2)) = x1 + 2·x2
POL(s(x1)) = 1 + x1
(15) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(16) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(17) TRUE
(18) 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:
f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))
The set Q consists of the following terms:
f(c(s(x0), x1))
g(c(x0, s(x1)))
We have to consider all minimal (P,Q,R)-chains.
(19) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(c(s(x), y)) → F(c(x, s(y)))
R is empty.
The set Q consists of the following terms:
f(c(s(x0), x1))
g(c(x0, s(x1)))
We have to consider all minimal (P,Q,R)-chains.
(21) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
f(c(s(x0), x1))
g(c(x0, s(x1)))
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(c(s(x), y)) → F(c(x, s(y)))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(23) 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 dependency pairs:
F(c(s(x), y)) → F(c(x, s(y)))
Used ordering: Polynomial interpretation [POLO]:
POL(F(x1)) = 2·x1
POL(c(x1, x2)) = 2·x1 + x2
POL(s(x1)) = 1 + x1
(24) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(25) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(26) TRUE