(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
C(x) → c(x)
c(c(x)) → x
b(b(x)) → B(x)
B(B(x)) → b(x)
c(B(c(b(c(x))))) → B(c(b(c(B(c(b(x)))))))
b(B(x)) → x
B(b(x)) → x
c(C(x)) → x
C(c(x)) → 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:
C(x) → c(x)
c(c(x)) → x
b(b(x)) → B(x)
B(B(x)) → b(x)
c(b(c(B(c(x))))) → b(c(B(c(b(c(B(x)))))))
B(b(x)) → x
b(B(x)) → x
C(c(x)) → x
c(C(x)) → x
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(B(x1)) = x1
POL(C(x1)) = 1 + x1
POL(b(x1)) = x1
POL(c(x1)) = x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
C(x) → c(x)
C(c(x)) → x
c(C(x)) → x
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
c(c(x)) → x
b(b(x)) → B(x)
B(B(x)) → b(x)
c(b(c(B(c(x))))) → b(c(B(c(b(c(B(x)))))))
B(b(x)) → x
b(B(x)) → x
Q is empty.
(5) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(B(x1)) = x1
POL(b(x1)) = x1
POL(c(x1)) = 1 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
c(c(x)) → x
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(x)) → B(x)
B(B(x)) → b(x)
c(b(c(B(c(x))))) → b(c(B(c(b(c(B(x)))))))
B(b(x)) → x
b(B(x)) → x
Q is empty.
(7) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B1(b(x)) → B2(x)
B2(B(x)) → B1(x)
C(b(c(B(c(x))))) → B1(c(B(c(b(c(B(x)))))))
C(b(c(B(c(x))))) → C(B(c(b(c(B(x))))))
C(b(c(B(c(x))))) → B2(c(b(c(B(x)))))
C(b(c(B(c(x))))) → C(b(c(B(x))))
C(b(c(B(c(x))))) → B1(c(B(x)))
C(b(c(B(c(x))))) → C(B(x))
C(b(c(B(c(x))))) → B2(x)
The TRS R consists of the following rules:
b(b(x)) → B(x)
B(B(x)) → b(x)
c(b(c(B(c(x))))) → b(c(B(c(b(c(B(x)))))))
B(b(x)) → x
b(B(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.
(10) Complex Obligation (AND)
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B2(B(x)) → B1(x)
B1(b(x)) → B2(x)
The TRS R consists of the following rules:
b(b(x)) → B(x)
B(B(x)) → b(x)
c(b(c(B(c(x))))) → b(c(B(c(b(c(B(x)))))))
B(b(x)) → x
b(B(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) 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.
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B2(B(x)) → B1(x)
B1(b(x)) → B2(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(14) 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:
- B1(b(x)) → B2(x)
The graph contains the following edges 1 > 1
- B2(B(x)) → B1(x)
The graph contains the following edges 1 > 1
(15) YES
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(b(c(B(c(x))))) → C(b(c(B(x))))
C(b(c(B(c(x))))) → C(B(c(b(c(B(x))))))
C(b(c(B(c(x))))) → C(B(x))
The TRS R consists of the following rules:
b(b(x)) → B(x)
B(B(x)) → b(x)
c(b(c(B(c(x))))) → b(c(B(c(b(c(B(x)))))))
B(b(x)) → x
b(B(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
C(b(c(B(c(x))))) → C(b(c(B(x))))
C(b(c(B(c(x))))) → C(B(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(B(x1)) = x1
POL(C(x1)) = x1
POL(b(x1)) = x1
POL(c(x1)) = 1 + x1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
B(B(x)) → b(x)
b(b(x)) → B(x)
B(b(x)) → x
c(b(c(B(c(x))))) → b(c(B(c(b(c(B(x)))))))
b(B(x)) → x
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(b(c(B(c(x))))) → C(B(c(b(c(B(x))))))
The TRS R consists of the following rules:
b(b(x)) → B(x)
B(B(x)) → b(x)
c(b(c(B(c(x))))) → b(c(B(c(b(c(B(x)))))))
B(b(x)) → x
b(B(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(19) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
C(b(c(B(c(x))))) → C(B(c(b(c(B(x))))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(b(x1)) = | | + | | / | -I | 0A | -I | \ |
| | | 0A | 0A | -I | | |
| \ | 0A | 0A | 0A | / |
| · | x1 |
| POL(c(x1)) = | | + | | / | 0A | -I | -I | \ |
| | | -I | -I | -I | | |
| \ | 0A | 0A | -I | / |
| · | x1 |
| POL(B(x1)) = | | + | | / | -I | 0A | -I | \ |
| | | 0A | 0A | -I | | |
| \ | 0A | 0A | 0A | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
B(B(x)) → b(x)
b(b(x)) → B(x)
B(b(x)) → x
c(b(c(B(c(x))))) → b(c(B(c(b(c(B(x)))))))
b(B(x)) → x
(20) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
b(b(x)) → B(x)
B(B(x)) → b(x)
c(b(c(B(c(x))))) → b(c(B(c(b(c(B(x)))))))
B(b(x)) → x
b(B(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(22) YES