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:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
id(x) → f(x, s(0))
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
id(x) → f(x, s(0))
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
id(x) → f(x, s(0))
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
id(x) → f(x, s(0))
Used ordering:
Polynomial interpretation [25]:
POL(0) = 0
POL(d) = 0
POL(double(x1)) = x1
POL(f(x1, x2)) = 2·x1 + 2·x2
POL(g(x1, x2)) = 2·x1 + x2
POL(h) = 0
POL(half(x1)) = x1
POL(id(x1)) = 2 + 2·x1
POL(s(x1)) = x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → 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(s(0), y) → y
Used ordering:
Polynomial interpretation [25]:
POL(0) = 0
POL(d) = 0
POL(double(x1)) = x1
POL(f(x1, x2)) = 2 + 2·x1 + 2·x2
POL(g(x1, x2)) = x1 + x2
POL(h) = 0
POL(half(x1)) = x1
POL(s(x1)) = x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
Q is empty.
We have applied [19,8] to switch to innermost. The TRS R 1 is
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
The TRS R 2 is
f(s(x), y) → f(half(s(x)), double(y))
The signature Sigma is {f}
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), 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:
HALF(x) → G(h, x)
F(s(x), y) → HALF(s(x))
DOUBLE(x) → G(d, x)
F(s(x), y) → DOUBLE(y)
F(s(x), y) → F(half(s(x)), double(y))
G(h, s(s(x))) → G(h, x)
G(d, s(x)) → G(d, x)
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
HALF(x) → G(h, x)
F(s(x), y) → HALF(s(x))
DOUBLE(x) → G(d, x)
F(s(x), y) → DOUBLE(y)
F(s(x), y) → F(half(s(x)), double(y))
G(h, s(s(x))) → G(h, x)
G(d, s(x)) → G(d, x)
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 4 less nodes.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G(h, s(s(x))) → G(h, x)
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), 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
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G(h, s(s(x))) → G(h, x)
R is empty.
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), 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.
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G(h, s(s(x))) → G(h, x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs:
- G(h, s(s(x))) → G(h, x)
The graph contains the following edges 1 >= 1, 2 > 2
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G(d, s(x)) → G(d, x)
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), 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
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G(d, s(x)) → G(d, x)
R is empty.
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), 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.
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), x1)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G(d, s(x)) → G(d, x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs:
- G(d, s(x)) → G(d, x)
The graph contains the following edges 1 >= 1, 2 > 2
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(s(x), y) → F(half(s(x)), double(y))
The TRS R consists of the following rules:
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), 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
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
F(s(x), y) → F(half(s(x)), double(y))
The TRS R consists of the following rules:
half(x) → g(h, x)
double(x) → g(d, x)
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
f(s(x0), 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(s(x0), x1)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
F(s(x), y) → F(half(s(x)), double(y))
The TRS R consists of the following rules:
half(x) → g(h, x)
double(x) → g(d, x)
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(s(x), y) → F(half(s(x)), double(y)) at position [0] we obtained the following new rules:
F(s(x), y) → F(g(h, s(x)), double(y))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(s(x), y) → F(g(h, s(x)), double(y))
The TRS R consists of the following rules:
half(x) → g(h, x)
double(x) → g(d, x)
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
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
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
F(s(x), y) → F(g(h, s(x)), double(y))
The TRS R consists of the following rules:
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
half(x0)
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.
half(x0)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
F(s(x), y) → F(g(h, s(x)), double(y))
The TRS R consists of the following rules:
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(s(x), y) → F(g(h, s(x)), double(y)) at position [1] we obtained the following new rules:
F(s(x), y) → F(g(h, s(x)), g(d, y))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(s(x), y) → F(g(h, s(x)), g(d, y))
The TRS R consists of the following rules:
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
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
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
F(s(x), y) → F(g(h, s(x)), g(d, y))
The TRS R consists of the following rules:
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
double(x0)
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.
double(x0)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
F(s(x), y) → F(g(h, s(x)), g(d, y))
The TRS R consists of the following rules:
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
F(s(x), y) → F(g(h, s(x)), g(d, y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( g(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
| M( F(x1, x2) ) = | 0 | + | | · | x1 | + | | · | x2 |
Matrix type:
We used a basic matrix type which is not further parametrizeable.
As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:
g(x, 0) → 0
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
The set Q consists of the following terms:
g(x0, 0)
g(d, s(x0))
g(h, s(0))
g(h, s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.