(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
p(0(x)) → s(s(0(s(s(p(x))))))
p(s(0(x))) → 0(x)
p(s(s(x))) → s(p(s(x)))
f(s(x)) → g(q(i(x)))
g(x) → f(p(p(x)))
q(i(x)) → q(s(x))
q(s(x)) → s(s(x))
i(x) → s(x)
Q is empty.
(1) QTRS Reverse (EQUIVALENT transformation)
We applied the QTRS Reverse Processor [REVERSE].
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
0(p(x)) → p(s(s(0(s(s(x))))))
0(s(p(x))) → 0(x)
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
g(x) → p(p(f(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)
Q is empty.
(3) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
0(p(x)) → p(s(s(0(s(s(x))))))
0(s(p(x))) → 0(x)
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
g(x) → p(p(f(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)
The set Q consists of the following terms:
0(p(x0))
0(s(p(x0)))
s(s(p(x0)))
s(f(x0))
g(x0)
s(q(x0))
i(x0)
(5) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
01(p(x)) → S(s(0(s(s(x)))))
01(p(x)) → S(0(s(s(x))))
01(p(x)) → 01(s(s(x)))
01(p(x)) → S(s(x))
01(p(x)) → S(x)
01(s(p(x))) → 01(x)
S(s(p(x))) → S(p(s(x)))
S(s(p(x))) → S(x)
S(f(x)) → I(q(g(x)))
S(f(x)) → G(x)
I(q(x)) → S(q(x))
S(q(x)) → S(s(x))
S(q(x)) → S(x)
I(x) → S(x)
The TRS R consists of the following rules:
0(p(x)) → p(s(s(0(s(s(x))))))
0(s(p(x))) → 0(x)
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
g(x) → p(p(f(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)
The set Q consists of the following terms:
0(p(x0))
0(s(p(x0)))
s(s(p(x0)))
s(f(x0))
g(x0)
s(q(x0))
i(x0)
We have to consider all minimal (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 6 less nodes.
(8) Complex Obligation (AND)
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(f(x)) → I(q(g(x)))
I(q(x)) → S(q(x))
S(q(x)) → S(s(x))
S(s(p(x))) → S(x)
S(q(x)) → S(x)
I(x) → S(x)
The TRS R consists of the following rules:
0(p(x)) → p(s(s(0(s(s(x))))))
0(s(p(x))) → 0(x)
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
g(x) → p(p(f(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)
The set Q consists of the following terms:
0(p(x0))
0(s(p(x0)))
s(s(p(x0)))
s(f(x0))
g(x0)
s(q(x0))
i(x0)
We have to consider all minimal (P,Q,R)-chains.
(10) 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.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(f(x)) → I(q(g(x)))
I(q(x)) → S(q(x))
S(q(x)) → S(s(x))
S(s(p(x))) → S(x)
S(q(x)) → S(x)
I(x) → S(x)
The TRS R consists of the following rules:
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)
g(x) → p(p(f(x)))
The set Q consists of the following terms:
0(p(x0))
0(s(p(x0)))
s(s(p(x0)))
s(f(x0))
g(x0)
s(q(x0))
i(x0)
We have to consider all minimal (P,Q,R)-chains.
(12) 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].
0(p(x0))
0(s(p(x0)))
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(f(x)) → I(q(g(x)))
I(q(x)) → S(q(x))
S(q(x)) → S(s(x))
S(s(p(x))) → S(x)
S(q(x)) → S(x)
I(x) → S(x)
The TRS R consists of the following rules:
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)
g(x) → p(p(f(x)))
The set Q consists of the following terms:
s(s(p(x0)))
s(f(x0))
g(x0)
s(q(x0))
i(x0)
We have to consider all minimal (P,Q,R)-chains.
(14) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
S(q(x)) → 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 - 2} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
g(x) → p(p(f(x)))
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(f(x)) → I(q(g(x)))
I(q(x)) → S(q(x))
S(q(x)) → S(s(x))
S(s(p(x))) → S(x)
I(x) → S(x)
The TRS R consists of the following rules:
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)
g(x) → p(p(f(x)))
The set Q consists of the following terms:
s(s(p(x0)))
s(f(x0))
g(x0)
s(q(x0))
i(x0)
We have to consider all minimal (P,Q,R)-chains.
(16) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
S(s(p(x))) → 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:
g(x) → p(p(f(x)))
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)
(17) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(f(x)) → I(q(g(x)))
I(q(x)) → S(q(x))
S(q(x)) → S(s(x))
I(x) → S(x)
The TRS R consists of the following rules:
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)
g(x) → p(p(f(x)))
The set Q consists of the following terms:
s(s(p(x0)))
s(f(x0))
g(x0)
s(q(x0))
i(x0)
We have to consider all minimal (P,Q,R)-chains.
(18) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
(19) Obligation:
Q DP problem:
The TRS P consists of the following rules:
I(x) → S(x)
S(f(x)) → I(q(g(x)))
The TRS R consists of the following rules:
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)
g(x) → p(p(f(x)))
The set Q consists of the following terms:
s(s(p(x0)))
s(f(x0))
g(x0)
s(q(x0))
i(x0)
We have to consider all minimal (P,Q,R)-chains.
(20) 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.
(21) Obligation:
Q DP problem:
The TRS P consists of the following rules:
I(x) → S(x)
S(f(x)) → I(q(g(x)))
The TRS R consists of the following rules:
g(x) → p(p(f(x)))
The set Q consists of the following terms:
s(s(p(x0)))
s(f(x0))
g(x0)
s(q(x0))
i(x0)
We have to consider all minimal (P,Q,R)-chains.
(22) 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].
s(s(p(x0)))
s(f(x0))
s(q(x0))
i(x0)
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
I(x) → S(x)
S(f(x)) → I(q(g(x)))
The TRS R consists of the following rules:
g(x) → p(p(f(x)))
The set Q consists of the following terms:
g(x0)
We have to consider all minimal (P,Q,R)-chains.
(24) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
I(x) → S(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(I(x1)) = 4 + 4·x1
POL(S(x1)) = 1 + 3·x1
POL(f(x1)) = 5 + 2·x1
POL(g(x1)) = 2 + 2·x1
POL(p(x1)) = 0
POL(q(x1)) = 3
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(f(x)) → I(q(g(x)))
The TRS R consists of the following rules:
g(x) → p(p(f(x)))
The set Q consists of the following terms:
g(x0)
We have to consider all minimal (P,Q,R)-chains.
(26) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(27) TRUE
(28) Obligation:
Q DP problem:
The TRS P consists of the following rules:
01(s(p(x))) → 01(x)
01(p(x)) → 01(s(s(x)))
The TRS R consists of the following rules:
0(p(x)) → p(s(s(0(s(s(x))))))
0(s(p(x))) → 0(x)
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
g(x) → p(p(f(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)
The set Q consists of the following terms:
0(p(x0))
0(s(p(x0)))
s(s(p(x0)))
s(f(x0))
g(x0)
s(q(x0))
i(x0)
We have to consider all minimal (P,Q,R)-chains.
(29) 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.
(30) Obligation:
Q DP problem:
The TRS P consists of the following rules:
01(s(p(x))) → 01(x)
01(p(x)) → 01(s(s(x)))
The TRS R consists of the following rules:
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)
g(x) → p(p(f(x)))
The set Q consists of the following terms:
0(p(x0))
0(s(p(x0)))
s(s(p(x0)))
s(f(x0))
g(x0)
s(q(x0))
i(x0)
We have to consider all minimal (P,Q,R)-chains.
(31) 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].
0(p(x0))
0(s(p(x0)))
(32) Obligation:
Q DP problem:
The TRS P consists of the following rules:
01(s(p(x))) → 01(x)
01(p(x)) → 01(s(s(x)))
The TRS R consists of the following rules:
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)
g(x) → p(p(f(x)))
The set Q consists of the following terms:
s(s(p(x0)))
s(f(x0))
g(x0)
s(q(x0))
i(x0)
We have to consider all minimal (P,Q,R)-chains.
(33) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
01(p(x)) → 01(s(s(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(s(x1)) = | | + | | / | -I | 0A | -I | \ |
| | | -I | 0A | -I | | |
| \ | 0A | 0A | -I | / |
| · | x1 |
| POL(p(x1)) = | | + | | / | 0A | 0A | 0A | \ |
| | | 0A | 0A | -I | | |
| \ | 0A | 1A | 0A | / |
| · | x1 |
| POL(f(x1)) = | | + | | / | 0A | -I | -I | \ |
| | | 0A | -I | -I | | |
| \ | 0A | -I | -I | / |
| · | x1 |
| POL(i(x1)) = | | + | | / | 0A | 0A | 0A | \ |
| | | 0A | 0A | 0A | | |
| \ | 0A | 0A | 0A | / |
| · | x1 |
| POL(q(x1)) = | | + | | / | -I | -I | -I | \ |
| | | -I | 0A | -I | | |
| \ | -I | -I | -I | / |
| · | x1 |
| POL(g(x1)) = | | + | | / | 1A | 0A | 0A | \ |
| | | 0A | -I | -I | | |
| \ | 1A | 0A | 0A | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)
g(x) → p(p(f(x)))
(34) Obligation:
Q DP problem:
The TRS P consists of the following rules:
01(s(p(x))) → 01(x)
The TRS R consists of the following rules:
s(s(p(x))) → s(p(s(x)))
s(f(x)) → i(q(g(x)))
i(q(x)) → s(q(x))
s(q(x)) → s(s(x))
i(x) → s(x)
g(x) → p(p(f(x)))
The set Q consists of the following terms:
s(s(p(x0)))
s(f(x0))
g(x0)
s(q(x0))
i(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:
01(s(p(x))) → 01(x)
R is empty.
The set Q consists of the following terms:
s(s(p(x0)))
s(f(x0))
g(x0)
s(q(x0))
i(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].
g(x0)
i(x0)
(38) Obligation:
Q DP problem:
The TRS P consists of the following rules:
01(s(p(x))) → 01(x)
R is empty.
The set Q consists of the following terms:
s(s(p(x0)))
s(f(x0))
s(q(x0))
We have to consider all minimal (P,Q,R)-chains.
(39) 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:
- 01(s(p(x))) → 01(x)
The graph contains the following edges 1 > 1
(40) YES