(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(x, s(y)) → PLUS(x, y)
TIMES(s(x), y) → PLUS(times(x, y), y)
TIMES(s(x), y) → TIMES(x, y)
P(s(s(x))) → P(s(x))
FAC(s(x)) → TIMES(fac(p(s(x))), s(x))
FAC(s(x)) → FAC(p(s(x)))
FAC(s(x)) → P(s(x))
The TRS R consists of the following rules:
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 3 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
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(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(7) Obligation:
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(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))
The set Q consists of the following terms:
plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(8) 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.
(9) Obligation:
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(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(10) 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].
plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))
(11) Obligation:
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.
(12) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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
(13) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(x, s(y)) → PLUS(x, y)
The TRS R consists of the following rules:
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(x, s(y)) → PLUS(x, y)
The TRS R consists of the following rules:
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))
The set Q consists of the following terms:
plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(17) 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.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(x, s(y)) → PLUS(x, y)
R is empty.
The set Q consists of the following terms:
plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(19) 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].
plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(x, s(y)) → PLUS(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:
- PLUS(x, s(y)) → PLUS(x, y)
The graph contains the following edges 1 >= 1, 2 > 2
(22) YES
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TIMES(s(x), y) → TIMES(x, y)
The TRS R consists of the following rules:
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(24) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TIMES(s(x), y) → TIMES(x, y)
The TRS R consists of the following rules:
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))
The set Q consists of the following terms:
plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(26) 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.
(27) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TIMES(s(x), y) → TIMES(x, y)
R is empty.
The set Q consists of the following terms:
plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(28) 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].
plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))
(29) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TIMES(s(x), y) → TIMES(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(30) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:
- TIMES(s(x), y) → TIMES(x, y)
The graph contains the following edges 1 > 1, 2 >= 2
(31) YES
(32) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FAC(s(x)) → FAC(p(s(x)))
The TRS R consists of the following rules:
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(33) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(34) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FAC(s(x)) → FAC(p(s(x)))
The TRS R consists of the following rules:
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
times(0, y) → 0
times(x, 0) → 0
times(s(x), y) → plus(times(x, y), y)
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
fac(s(x)) → times(fac(p(s(x))), s(x))
The set Q consists of the following terms:
plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(35) 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.
(36) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FAC(s(x)) → FAC(p(s(x)))
The TRS R consists of the following rules:
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
The set Q consists of the following terms:
plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
p(s(s(x0)))
p(s(0))
fac(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(37) 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].
plus(x0, 0)
plus(x0, s(x1))
times(0, x0)
times(x0, 0)
times(s(x0), x1)
fac(s(x0))
(38) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FAC(s(x)) → FAC(p(s(x)))
The TRS R consists of the following rules:
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
The set Q consists of the following terms:
p(s(s(x0)))
p(s(0))
We have to consider all minimal (P,Q,R)-chains.
(39) 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 rules of the TRS R:
p(s(0)) → 0
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(FAC(x1)) = x1
POL(p(x1)) = x1
POL(s(x1)) = 2 + 2·x1
(40) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FAC(s(x)) → FAC(p(s(x)))
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(s(x0)))
p(s(0))
We have to consider all minimal (P,Q,R)-chains.
(41) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
FAC(s(x)) → FAC(p(s(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( p(x1) ) = max{0, x1 - 1} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
p(s(s(x))) → s(p(s(x)))
(42) Obligation:
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(s(x0)))
p(s(0))
We have to consider all minimal (P,Q,R)-chains.
(43) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(44) YES