(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → b(c(a(x)))
b(c(x)) → c(b(b(x)))
a(c(x)) → c(a(b(x)))
a(a(x)) → a(d(d(d(x))))
d(a(x)) → d(d(c(x)))
a(d(d(c(x)))) → a(a(a(d(x))))
e(e(f(f(x)))) → f(f(f(e(e(x)))))
e(x) → a(x)
b(d(x)) → d(d(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:
b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
c(a(x)) → b(a(c(x)))
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(d(d(a(x)))) → d(a(a(a(x))))
f(f(e(e(x)))) → e(e(f(f(f(x)))))
e(x) → a(x)
d(b(x)) → d(d(x))
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(a(x1)) = x1
POL(b(x1)) = x1
POL(c(x1)) = x1
POL(d(x1)) = x1
POL(e(x1)) = 1 + x1
POL(f(x1)) = x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
e(x) → a(x)
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
c(a(x)) → b(a(c(x)))
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(d(d(a(x)))) → d(a(a(a(x))))
f(f(e(e(x)))) → e(e(f(f(f(x)))))
d(b(x)) → d(d(x))
Q is empty.
(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:
B(a(x)) → A(c(b(x)))
B(a(x)) → C(b(x))
B(a(x)) → B(x)
C(b(x)) → B(b(c(x)))
C(b(x)) → B(c(x))
C(b(x)) → C(x)
C(a(x)) → B(a(c(x)))
C(a(x)) → A(c(x))
C(a(x)) → C(x)
A(a(x)) → D(d(d(a(x))))
A(a(x)) → D(d(a(x)))
A(a(x)) → D(a(x))
A(d(x)) → C(d(d(x)))
A(d(x)) → D(d(x))
C(d(d(a(x)))) → D(a(a(a(x))))
C(d(d(a(x)))) → A(a(a(x)))
C(d(d(a(x)))) → A(a(x))
F(f(e(e(x)))) → F(f(f(x)))
F(f(e(e(x)))) → F(f(x))
F(f(e(e(x)))) → F(x)
D(b(x)) → D(d(x))
D(b(x)) → D(x)
The TRS R consists of the following rules:
b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
c(a(x)) → b(a(c(x)))
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(d(d(a(x)))) → d(a(a(a(x))))
f(f(e(e(x)))) → e(e(f(f(f(x)))))
d(b(x)) → d(d(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 5 less nodes.
(8) Complex Obligation (AND)
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(b(x)) → D(x)
D(b(x)) → D(d(x))
The TRS R consists of the following rules:
b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
c(a(x)) → b(a(c(x)))
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(d(d(a(x)))) → d(a(a(a(x))))
f(f(e(e(x)))) → e(e(f(f(f(x)))))
d(b(x)) → d(d(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(b(x)) → D(x)
D(b(x)) → D(d(x))
The TRS R consists of the following rules:
d(b(x)) → d(d(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(b(x)) → D(x)
The TRS R consists of the following rules:
d(b(x)) → d(d(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(14) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(b(x)) → D(x)
The TRS R consists of the following rules:
d(b(x)) → d(d(x))
The set Q consists of the following terms:
d(b(x0))
We have to consider all minimal (P,Q,R)-chains.
(16) 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.
(17) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(b(x)) → D(x)
R is empty.
The set Q consists of the following terms:
d(b(x0))
We have to consider all minimal (P,Q,R)-chains.
(18) 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].
d(b(x0))
(19) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(b(x)) → D(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(20) 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:
- D(b(x)) → D(x)
The graph contains the following edges 1 > 1
(21) YES
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(f(e(e(x)))) → F(f(x))
F(f(e(e(x)))) → F(f(f(x)))
F(f(e(e(x)))) → F(x)
The TRS R consists of the following rules:
b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
c(a(x)) → b(a(c(x)))
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(d(d(a(x)))) → d(a(a(a(x))))
f(f(e(e(x)))) → e(e(f(f(f(x)))))
d(b(x)) → d(d(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(23) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(f(e(e(x)))) → F(f(x))
F(f(e(e(x)))) → F(f(f(x)))
F(f(e(e(x)))) → F(x)
The TRS R consists of the following rules:
f(f(e(e(x)))) → e(e(f(f(f(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(25) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(26) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(f(e(e(x)))) → F(f(x))
F(f(e(e(x)))) → F(f(f(x)))
F(f(e(e(x)))) → F(x)
The TRS R consists of the following rules:
f(f(e(e(x)))) → e(e(f(f(f(x)))))
The set Q consists of the following terms:
f(f(e(e(x0))))
We have to consider all minimal (P,Q,R)-chains.
(27) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
F(f(e(e(x)))) → F(f(x))
F(f(e(e(x)))) → F(f(f(x)))
F(f(e(e(x)))) → F(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(F(x1)) = x1
POL(e(x1)) = 1 + x1
POL(f(x1)) = x1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
f(f(e(e(x)))) → e(e(f(f(f(x)))))
(28) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(f(e(e(x)))) → e(e(f(f(f(x)))))
The set Q consists of the following terms:
f(f(e(e(x0))))
We have to consider all minimal (P,Q,R)-chains.
(29) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(30) YES
(31) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(d(x)) → C(d(d(x)))
C(b(x)) → B(b(c(x)))
B(a(x)) → A(c(b(x)))
B(a(x)) → C(b(x))
C(b(x)) → B(c(x))
B(a(x)) → B(x)
C(b(x)) → C(x)
C(a(x)) → B(a(c(x)))
C(a(x)) → A(c(x))
C(a(x)) → C(x)
C(d(d(a(x)))) → A(a(a(x)))
C(d(d(a(x)))) → A(a(x))
The TRS R consists of the following rules:
b(a(x)) → a(c(b(x)))
c(b(x)) → b(b(c(x)))
c(a(x)) → b(a(c(x)))
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(d(d(a(x)))) → d(a(a(a(x))))
f(f(e(e(x)))) → e(e(f(f(f(x)))))
d(b(x)) → d(d(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(32) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(33) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(d(x)) → C(d(d(x)))
C(b(x)) → B(b(c(x)))
B(a(x)) → A(c(b(x)))
B(a(x)) → C(b(x))
C(b(x)) → B(c(x))
B(a(x)) → B(x)
C(b(x)) → C(x)
C(a(x)) → B(a(c(x)))
C(a(x)) → A(c(x))
C(a(x)) → C(x)
C(d(d(a(x)))) → A(a(a(x)))
C(d(d(a(x)))) → A(a(x))
The TRS R consists of the following rules:
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(b(x)) → b(b(c(x)))
b(a(x)) → a(c(b(x)))
c(a(x)) → b(a(c(x)))
d(b(x)) → d(d(x))
c(d(d(a(x)))) → d(a(a(a(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(34) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
B(a(x)) → A(c(b(x)))
B(a(x)) → C(b(x))
B(a(x)) → B(x)
C(a(x)) → A(c(x))
C(a(x)) → C(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = 0
POL(B(x1)) = x1
POL(C(x1)) = x1
POL(a(x1)) = 1 + x1
POL(b(x1)) = x1
POL(c(x1)) = x1
POL(d(x1)) = 0
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
d(b(x)) → d(d(x))
a(d(x)) → c(d(d(x)))
c(b(x)) → b(b(c(x)))
b(a(x)) → a(c(b(x)))
c(a(x)) → b(a(c(x)))
c(d(d(a(x)))) → d(a(a(a(x))))
a(a(x)) → d(d(d(a(x))))
(35) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(d(x)) → C(d(d(x)))
C(b(x)) → B(b(c(x)))
C(b(x)) → B(c(x))
C(b(x)) → C(x)
C(a(x)) → B(a(c(x)))
C(d(d(a(x)))) → A(a(a(x)))
C(d(d(a(x)))) → A(a(x))
The TRS R consists of the following rules:
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(b(x)) → b(b(c(x)))
b(a(x)) → a(c(b(x)))
c(a(x)) → b(a(c(x)))
d(b(x)) → d(d(x))
c(d(d(a(x)))) → d(a(a(a(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(36) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
(37) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(b(x)) → C(x)
C(d(d(a(x)))) → A(a(a(x)))
A(d(x)) → C(d(d(x)))
C(d(d(a(x)))) → A(a(x))
The TRS R consists of the following rules:
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(b(x)) → b(b(c(x)))
b(a(x)) → a(c(b(x)))
c(a(x)) → b(a(c(x)))
d(b(x)) → d(d(x))
c(d(d(a(x)))) → d(a(a(a(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(38) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
C(b(x)) → C(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = 0
POL(C(x1)) = x1
POL(a(x1)) = 0
POL(b(x1)) = 1 + x1
POL(c(x1)) = 0
POL(d(x1)) = 0
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
d(b(x)) → d(d(x))
(39) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(d(d(a(x)))) → A(a(a(x)))
A(d(x)) → C(d(d(x)))
C(d(d(a(x)))) → A(a(x))
The TRS R consists of the following rules:
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(b(x)) → b(b(c(x)))
b(a(x)) → a(c(b(x)))
c(a(x)) → b(a(c(x)))
d(b(x)) → d(d(x))
c(d(d(a(x)))) → d(a(a(a(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(40) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
C(d(d(a(x)))) → A(a(a(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(d(x1)) = | | + | | / | -I | 0A | 0A | \ |
| | | -I | -I | 0A | | |
| \ | -I | -I | -I | / |
| · | x1 |
| POL(a(x1)) = | | + | | / | -I | 0A | -I | \ |
| | | 0A | -I | -I | | |
| \ | 1A | 0A | -I | / |
| · | x1 |
| POL(c(x1)) = | | + | | / | 0A | 0A | -I | \ |
| | | 0A | 0A | -I | | |
| \ | 0A | 0A | 1A | / |
| · | x1 |
| POL(b(x1)) = | | + | | / | 0A | 0A | -I | \ |
| | | 0A | 0A | -I | | |
| \ | 1A | 0A | 0A | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(b(x)) → b(b(c(x)))
b(a(x)) → a(c(b(x)))
c(a(x)) → b(a(c(x)))
d(b(x)) → d(d(x))
c(d(d(a(x)))) → d(a(a(a(x))))
(41) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(d(x)) → C(d(d(x)))
C(d(d(a(x)))) → A(a(x))
The TRS R consists of the following rules:
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(b(x)) → b(b(c(x)))
b(a(x)) → a(c(b(x)))
c(a(x)) → b(a(c(x)))
d(b(x)) → d(d(x))
c(d(d(a(x)))) → d(a(a(a(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(42) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
A(d(x)) → C(d(d(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(d(x1)) = | | + | | / | -I | -I | -I | \ |
| | | 0A | -I | -I | | |
| \ | 0A | 0A | -I | / |
| · | x1 |
| POL(a(x1)) = | | + | | / | 0A | 1A | 1A | \ |
| | | -I | 0A | 0A | | |
| \ | -I | 0A | 0A | / |
| · | x1 |
| POL(b(x1)) = | | + | | / | 0A | 0A | 0A | \ |
| | | -I | 0A | 0A | | |
| \ | -I | 0A | -I | / |
| · | x1 |
| POL(c(x1)) = | | + | | / | 0A | 0A | -I | \ |
| | | -I | 0A | 0A | | |
| \ | -I | 0A | 0A | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
d(b(x)) → d(d(x))
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(b(x)) → b(b(c(x)))
b(a(x)) → a(c(b(x)))
c(a(x)) → b(a(c(x)))
c(d(d(a(x)))) → d(a(a(a(x))))
(43) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(d(d(a(x)))) → A(a(x))
The TRS R consists of the following rules:
a(a(x)) → d(d(d(a(x))))
a(d(x)) → c(d(d(x)))
c(b(x)) → b(b(c(x)))
b(a(x)) → a(c(b(x)))
c(a(x)) → b(a(c(x)))
d(b(x)) → d(d(x))
c(d(d(a(x)))) → d(a(a(a(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(44) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(45) TRUE