Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R 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))
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R 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))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R 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))
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
f(c(s(x), s(y))) → g(c(x, y))
g(c(s(x), s(y))) → f(c(x, y))
Used ordering:
Polynomial interpretation [25]:
POL(c(x1, x2)) = x1 + x2
POL(f(x1)) = x1
POL(g(x1)) = 1 + x1
POL(s(x1)) = 1 + x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
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.
We have applied [19,8] 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}
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
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)))
Using Dependency Pairs [1,15] we result in the following initial DP problem:
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.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
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.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
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.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
f(c(s(x0), x1))
g(c(x0, s(x1)))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
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.
By using the rule removal processor [15] with the following polynomial ordering [25], 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: POLO with Polynomial interpretation [25]:
POL(G(x1)) = 2·x1
POL(c(x1, x2)) = x1 + 2·x2
POL(s(x1)) = 1 + x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
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.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
f(c(s(x0), x1))
g(c(x0, s(x1)))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
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.
By using the rule removal processor [15] with the following polynomial ordering [25], 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: POLO with Polynomial interpretation [25]:
POL(F(x1)) = 2·x1
POL(c(x1, x2)) = 2·x1 + x2
POL(s(x1)) = 1 + x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.