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:
plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))
Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))
Q is empty.
The TRS is overlay and locally confluent. By [19] we can switch to innermost.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))
The set Q consists of the following terms:
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
FAC(s(x), y) → TIMES(s(x), y)
FAC(s(x), y) → P(s(x))
TIMES(s(x), y) → P(s(x))
TIMES(s(x), y) → PLUS(y, times(p(s(x)), y))
FAC(s(x), y) → FAC(p(s(x)), times(s(x), y))
P(s(s(x))) → P(s(x))
PLUS(s(x), y) → PLUS(p(s(x)), y)
FACTORIAL(x) → FAC(x, s(0))
PLUS(s(x), y) → P(s(x))
TIMES(s(x), y) → TIMES(p(s(x)), y)
The TRS R consists of the following rules:
plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))
The set Q consists of the following terms:
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
FAC(s(x), y) → TIMES(s(x), y)
FAC(s(x), y) → P(s(x))
TIMES(s(x), y) → P(s(x))
TIMES(s(x), y) → PLUS(y, times(p(s(x)), y))
FAC(s(x), y) → FAC(p(s(x)), times(s(x), y))
P(s(s(x))) → P(s(x))
PLUS(s(x), y) → PLUS(p(s(x)), y)
FACTORIAL(x) → FAC(x, s(0))
PLUS(s(x), y) → P(s(x))
TIMES(s(x), y) → TIMES(p(s(x)), y)
The TRS R consists of the following rules:
plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))
The set Q consists of the following terms:
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 6 less nodes.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
P(s(s(x))) → P(s(x))
The TRS R consists of the following rules:
plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))
The set Q consists of the following terms:
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(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
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
P(s(s(x))) → P(s(x))
R is empty.
The set Q consists of the following terms:
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(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.
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(x0)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
P(s(s(x))) → P(s(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:
- P(s(s(x))) → P(s(x))
The graph contains the following edges 1 > 1
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PLUS(s(x), y) → PLUS(p(s(x)), y)
The TRS R consists of the following rules:
plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))
The set Q consists of the following terms:
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(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
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PLUS(s(x), y) → PLUS(p(s(x)), y)
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(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.
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
fac(0, x0)
fac(s(x0), x1)
factorial(x0)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PLUS(s(x), y) → PLUS(p(s(x)), y)
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(s(0))
p(s(s(x0)))
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 rules of the TRS R:
p(s(0)) → 0
Used ordering: POLO with Polynomial interpretation [25]:
POL(0) = 1
POL(PLUS(x1, x2)) = x1 + x2
POL(p(x1)) = x1
POL(s(x1)) = 2·x1
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PLUS(s(x), y) → PLUS(p(s(x)), y)
The TRS R consists of the following rules:
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(s(0))
p(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.
PLUS(s(x), y) → PLUS(p(s(x)), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
Tuple symbols:
| M( PLUS(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:
p(s(s(x))) → s(p(s(x)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(s(0))
p(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.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TIMES(s(x), y) → TIMES(p(s(x)), y)
The TRS R consists of the following rules:
plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))
The set Q consists of the following terms:
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(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
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TIMES(s(x), y) → TIMES(p(s(x)), y)
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(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.
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
fac(0, x0)
fac(s(x0), x1)
factorial(x0)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TIMES(s(x), y) → TIMES(p(s(x)), y)
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(s(0))
p(s(s(x0)))
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 rules of the TRS R:
p(s(0)) → 0
Used ordering: POLO with Polynomial interpretation [25]:
POL(0) = 1
POL(TIMES(x1, x2)) = x1 + x2
POL(p(x1)) = x1
POL(s(x1)) = 2·x1
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TIMES(s(x), y) → TIMES(p(s(x)), y)
The TRS R consists of the following rules:
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(s(0))
p(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.
TIMES(s(x), y) → TIMES(p(s(x)), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
Tuple symbols:
| M( TIMES(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:
p(s(s(x))) → s(p(s(x)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
p(s(s(x))) → s(p(s(x)))
The set Q consists of the following terms:
p(s(0))
p(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.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
FAC(s(x), y) → FAC(p(s(x)), times(s(x), y))
The TRS R consists of the following rules:
plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))
The set Q consists of the following terms:
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(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
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
FAC(s(x), y) → FAC(p(s(x)), times(s(x), y))
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
times(s(x), y) → plus(y, times(p(s(x)), y))
times(0, y) → 0
plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
The set Q consists of the following terms:
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
fac(0, x0)
fac(s(x0), x1)
factorial(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.
fac(0, x0)
fac(s(x0), x1)
factorial(x0)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
FAC(s(x), y) → FAC(p(s(x)), times(s(x), y))
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
times(s(x), y) → plus(y, times(p(s(x)), y))
times(0, y) → 0
plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
The set Q consists of the following terms:
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(s(s(x0)))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule FAC(s(x), y) → FAC(p(s(x)), times(s(x), y)) at position [1] we obtained the following new rules:
FAC(s(x), y) → FAC(p(s(x)), plus(y, times(p(s(x)), y)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
FAC(s(x), y) → FAC(p(s(x)), plus(y, times(p(s(x)), y)))
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
times(s(x), y) → plus(y, times(p(s(x)), y))
times(0, y) → 0
plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
The set Q consists of the following terms:
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(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.
FAC(s(x), y) → FAC(p(s(x)), plus(y, times(p(s(x)), y)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( plus(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
| M( times(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
| M( FAC(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:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
times(s(x), y) → plus(y, times(p(s(x)), y))
times(0, y) → 0
plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
The set Q consists of the following terms:
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
p(s(0))
p(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.