(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(a(x))) → b(c(x))
b(b(b(x))) → c(b(x))
c(x) → a(b(x))
c(d(x)) → d(c(b(a(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:
a(b(a(x))) → c(b(x))
b(b(b(x))) → b(c(x))
c(x) → b(a(x))
d(c(x)) → a(b(c(d(x))))
Q is empty.
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(a(x))) → C(b(x))
A(b(a(x))) → B(x)
B(b(b(x))) → B(c(x))
B(b(b(x))) → C(x)
C(x) → B(a(x))
C(x) → A(x)
D(c(x)) → A(b(c(d(x))))
D(c(x)) → B(c(d(x)))
D(c(x)) → C(d(x))
D(c(x)) → D(x)
The TRS R consists of the following rules:
a(b(a(x))) → c(b(x))
b(b(b(x))) → b(c(x))
c(x) → b(a(x))
d(c(x)) → a(b(c(d(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(x) → B(a(x))
B(b(b(x))) → B(c(x))
B(b(b(x))) → C(x)
C(x) → A(x)
A(b(a(x))) → C(b(x))
A(b(a(x))) → B(x)
The TRS R consists of the following rules:
a(b(a(x))) → c(b(x))
b(b(b(x))) → b(c(x))
c(x) → b(a(x))
d(c(x)) → a(b(c(d(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(x) → B(a(x))
B(b(b(x))) → B(c(x))
B(b(b(x))) → C(x)
C(x) → A(x)
A(b(a(x))) → C(b(x))
A(b(a(x))) → B(x)
The TRS R consists of the following rules:
b(b(b(x))) → b(c(x))
c(x) → b(a(x))
a(b(a(x))) → c(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) 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 dependency pairs:
C(x) → B(a(x))
A(b(a(x))) → C(b(x))
A(b(a(x))) → B(x)
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = 2 + x1
POL(B(x1)) = x1
POL(C(x1)) = 2 + x1
POL(a(x1)) = 1 + x1
POL(b(x1)) = 1 + x1
POL(c(x1)) = 2 + x1
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(b(b(x))) → B(c(x))
B(b(b(x))) → C(x)
C(x) → A(x)
The TRS R consists of the following rules:
b(b(b(x))) → b(c(x))
c(x) → b(a(x))
a(b(a(x))) → c(b(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 2 less nodes.
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(b(b(x))) → B(c(x))
The TRS R consists of the following rules:
b(b(b(x))) → b(c(x))
c(x) → b(a(x))
a(b(a(x))) → c(b(x))
Q is empty.
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.
B(b(b(x))) → B(c(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(b(x1)) = | | + | | / | 1A | -I | 1A | \ |
| | | -I | -I | -I | | |
| \ | -I | 0A | -I | / |
| · | x1 |
| POL(c(x1)) = | | + | | / | -I | 0A | 1A | \ |
| | | -I | -I | -I | | |
| \ | -I | 1A | 0A | / |
| · | x1 |
| POL(a(x1)) = | | + | | / | -I | -I | 0A | \ |
| | | -I | 1A | 0A | | |
| \ | -I | -I | 0A | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
c(x) → b(a(x))
a(b(a(x))) → c(b(x))
b(b(b(x))) → b(c(x))
(15) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
b(b(b(x))) → b(c(x))
c(x) → b(a(x))
a(b(a(x))) → c(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(16) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(17) YES
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(c(x)) → D(x)
The TRS R consists of the following rules:
a(b(a(x))) → c(b(x))
b(b(b(x))) → b(c(x))
c(x) → b(a(x))
d(c(x)) → a(b(c(d(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(19) 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.
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(c(x)) → D(x)
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:
- D(c(x)) → D(x)
The graph contains the following edges 1 > 1
(22) YES