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(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))
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(nil) → nil
Used ordering:
Polynomial interpretation [25]:
POL(.(x1, x2)) = x1 + x2
POL(f(x1)) = 1 + x1
POL(g(x1)) = x1
POL(nil) = 0
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
g(nil) → nil
Used ordering:
Polynomial interpretation [25]:
POL(.(x1, x2)) = x1 + x2
POL(f(x1)) = x1
POL(g(x1)) = 2 + x1
POL(nil) = 0
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))
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(.(nil, y)) → .(nil, f(y))
g(.(x, nil)) → .(g(x), nil)
Used ordering:
Polynomial interpretation [25]:
POL(.(x1, x2)) = x1 + x2
POL(f(x1)) = 2·x1
POL(g(x1)) = 2·x1
POL(nil) = 2
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(.(x, .(y, z))) → g(.(.(x, y), z))
Q is empty.
We have applied [19,8] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(.(x, .(y, z))) → g(.(.(x, y), z))
The signature Sigma is {f, g}
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(.(x, .(y, z))) → g(.(.(x, y), z))
The set Q consists of the following terms:
f(.(.(x0, x1), x2))
g(.(x0, .(x1, x2)))
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(.(.(x, y), z)) → F(.(x, .(y, z)))
G(.(x, .(y, z))) → G(.(.(x, y), z))
The TRS R consists of the following rules:
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(.(x, .(y, z))) → g(.(.(x, y), z))
The set Q consists of the following terms:
f(.(.(x0, x1), x2))
g(.(x0, .(x1, x2)))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(.(.(x, y), z)) → F(.(x, .(y, z)))
G(.(x, .(y, z))) → G(.(.(x, y), z))
The TRS R consists of the following rules:
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(.(x, .(y, z))) → g(.(.(x, y), z))
The set Q consists of the following terms:
f(.(.(x0, x1), x2))
g(.(x0, .(x1, x2)))
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
↳ RRRPoloQTRSProof
↳ 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(.(x, .(y, z))) → G(.(.(x, y), z))
The TRS R consists of the following rules:
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(.(x, .(y, z))) → g(.(.(x, y), z))
The set Q consists of the following terms:
f(.(.(x0, x1), x2))
g(.(x0, .(x1, x2)))
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
↳ RRRPoloQTRSProof
↳ 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(.(x, .(y, z))) → G(.(.(x, y), z))
R is empty.
The set Q consists of the following terms:
f(.(.(x0, x1), x2))
g(.(x0, .(x1, x2)))
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(.(.(x0, x1), x2))
g(.(x0, .(x1, x2)))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ 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(.(x, .(y, z))) → G(.(.(x, y), z))
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(.(x, .(y, z))) → G(.(.(x, y), z))
Used ordering: POLO with Polynomial interpretation [25]:
POL(.(x1, x2)) = 2 + x1 + 2·x2
POL(G(x1)) = 2·x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ 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
↳ RRRPoloQTRSProof
↳ 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(.(.(x, y), z)) → F(.(x, .(y, z)))
The TRS R consists of the following rules:
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(.(x, .(y, z))) → g(.(.(x, y), z))
The set Q consists of the following terms:
f(.(.(x0, x1), x2))
g(.(x0, .(x1, x2)))
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
↳ RRRPoloQTRSProof
↳ 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(.(.(x, y), z)) → F(.(x, .(y, z)))
R is empty.
The set Q consists of the following terms:
f(.(.(x0, x1), x2))
g(.(x0, .(x1, x2)))
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(.(.(x0, x1), x2))
g(.(x0, .(x1, x2)))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ 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(.(.(x, y), z)) → F(.(x, .(y, z)))
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(.(.(x, y), z)) → F(.(x, .(y, z)))
Used ordering: POLO with Polynomial interpretation [25]:
POL(.(x1, x2)) = 2 + 2·x1 + x2
POL(F(x1)) = 2·x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ 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.