(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(a(x))) → a(b(c(x)))
c(a(x)) → a(c(x))
b(c(a(x))) → a(b(c(x)))
c(b(x)) → d(x)
a(d(x)) → d(a(x))
d(x) → b(a(x))
L(a(a(x))) → L(a(b(c(x))))
c(R(x)) → c(b(R(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:
B(a(a(x))) → A(b(c(x)))
B(a(a(x))) → B(c(x))
B(a(a(x))) → C(x)
C(a(x)) → A(c(x))
C(a(x)) → C(x)
B(c(a(x))) → A(b(c(x)))
B(c(a(x))) → B(c(x))
B(c(a(x))) → C(x)
C(b(x)) → D(x)
A(d(x)) → D(a(x))
A(d(x)) → A(x)
D(x) → B(a(x))
D(x) → A(x)
L1(a(a(x))) → L1(a(b(c(x))))
L1(a(a(x))) → A(b(c(x)))
L1(a(a(x))) → B(c(x))
L1(a(a(x))) → C(x)
C(R(x)) → C(b(R(x)))
C(R(x)) → B(R(x))
The TRS R consists of the following rules:
b(a(a(x))) → a(b(c(x)))
c(a(x)) → a(c(x))
b(c(a(x))) → a(b(c(x)))
c(b(x)) → d(x)
a(d(x)) → d(a(x))
d(x) → b(a(x))
L(a(a(x))) → L(a(b(c(x))))
c(R(x)) → c(b(R(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 2 SCCs with 4 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(d(x)) → D(a(x))
D(x) → B(a(x))
B(a(a(x))) → A(b(c(x)))
A(d(x)) → A(x)
B(a(a(x))) → B(c(x))
B(a(a(x))) → C(x)
C(a(x)) → A(c(x))
C(a(x)) → C(x)
C(b(x)) → D(x)
D(x) → A(x)
C(R(x)) → C(b(R(x)))
B(c(a(x))) → A(b(c(x)))
B(c(a(x))) → B(c(x))
B(c(a(x))) → C(x)
The TRS R consists of the following rules:
b(a(a(x))) → a(b(c(x)))
c(a(x)) → a(c(x))
b(c(a(x))) → a(b(c(x)))
c(b(x)) → d(x)
a(d(x)) → d(a(x))
d(x) → b(a(x))
L(a(a(x))) → L(a(b(c(x))))
c(R(x)) → c(b(R(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) 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.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(d(x)) → D(a(x))
D(x) → B(a(x))
B(a(a(x))) → A(b(c(x)))
A(d(x)) → A(x)
B(a(a(x))) → B(c(x))
B(a(a(x))) → C(x)
C(a(x)) → A(c(x))
C(a(x)) → C(x)
C(b(x)) → D(x)
D(x) → A(x)
C(R(x)) → C(b(R(x)))
B(c(a(x))) → A(b(c(x)))
B(c(a(x))) → B(c(x))
B(c(a(x))) → C(x)
The TRS R consists of the following rules:
c(a(x)) → a(c(x))
c(b(x)) → d(x)
c(R(x)) → c(b(R(x)))
a(d(x)) → d(a(x))
d(x) → b(a(x))
b(a(a(x))) → a(b(c(x)))
b(c(a(x))) → a(b(c(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
A(d(x)) → A(x)
B(a(a(x))) → B(c(x))
B(a(a(x))) → C(x)
C(a(x)) → C(x)
B(c(a(x))) → B(c(x))
B(c(a(x))) → C(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = 1 + x1
POL(B(x1)) = x1
POL(C(x1)) = 1 + x1
POL(D(x1)) = 1 + x1
POL(R(x1)) = 1 + x1
POL(a(x1)) = 1 + x1
POL(b(x1)) = x1
POL(c(x1)) = 1 + x1
POL(d(x1)) = 1 + x1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
d(x) → b(a(x))
b(a(a(x))) → a(b(c(x)))
a(d(x)) → d(a(x))
b(c(a(x))) → a(b(c(x)))
c(a(x)) → a(c(x))
c(b(x)) → d(x)
c(R(x)) → c(b(R(x)))
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(d(x)) → D(a(x))
D(x) → B(a(x))
B(a(a(x))) → A(b(c(x)))
C(a(x)) → A(c(x))
C(b(x)) → D(x)
D(x) → A(x)
C(R(x)) → C(b(R(x)))
B(c(a(x))) → A(b(c(x)))
The TRS R consists of the following rules:
c(a(x)) → a(c(x))
c(b(x)) → d(x)
c(R(x)) → c(b(R(x)))
a(d(x)) → d(a(x))
d(x) → b(a(x))
b(a(a(x))) → a(b(c(x)))
b(c(a(x))) → a(b(c(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(x) → B(a(x))
B(a(a(x))) → A(b(c(x)))
A(d(x)) → D(a(x))
D(x) → A(x)
B(c(a(x))) → A(b(c(x)))
The TRS R consists of the following rules:
c(a(x)) → a(c(x))
c(b(x)) → d(x)
c(R(x)) → c(b(R(x)))
a(d(x)) → d(a(x))
d(x) → b(a(x))
b(a(a(x))) → a(b(c(x)))
b(c(a(x))) → a(b(c(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
A(d(x)) → D(a(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = x1
POL(B(x1)) = 0
POL(D(x1)) = x1
POL(R(x1)) = 1 + x1
POL(a(x1)) = x1
POL(b(x1)) = 0
POL(c(x1)) = 0
POL(d(x1)) = 1 + x1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
d(x) → b(a(x))
b(a(a(x))) → a(b(c(x)))
a(d(x)) → d(a(x))
b(c(a(x))) → a(b(c(x)))
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(x) → B(a(x))
B(a(a(x))) → A(b(c(x)))
D(x) → A(x)
B(c(a(x))) → A(b(c(x)))
The TRS R consists of the following rules:
c(a(x)) → a(c(x))
c(b(x)) → d(x)
c(R(x)) → c(b(R(x)))
a(d(x)) → d(a(x))
d(x) → b(a(x))
b(a(a(x))) → a(b(c(x)))
b(c(a(x))) → a(b(c(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(14) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 4 less nodes.
(15) TRUE
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
L1(a(a(x))) → L1(a(b(c(x))))
The TRS R consists of the following rules:
b(a(a(x))) → a(b(c(x)))
c(a(x)) → a(c(x))
b(c(a(x))) → a(b(c(x)))
c(b(x)) → d(x)
a(d(x)) → d(a(x))
d(x) → b(a(x))
L(a(a(x))) → L(a(b(c(x))))
c(R(x)) → c(b(R(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) 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.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
L1(a(a(x))) → L1(a(b(c(x))))
The TRS R consists of the following rules:
c(a(x)) → a(c(x))
c(b(x)) → d(x)
c(R(x)) → c(b(R(x)))
a(d(x)) → d(a(x))
d(x) → b(a(x))
b(a(a(x))) → a(b(c(x)))
b(c(a(x))) → a(b(c(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.
L1(a(a(x))) → L1(a(b(c(x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( L1(x1) ) = max{0, x1 - 2} |
| POL( b(x1) ) = max{0, x1 - 2} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
c(a(x)) → a(c(x))
c(b(x)) → d(x)
c(R(x)) → c(b(R(x)))
d(x) → b(a(x))
b(a(a(x))) → a(b(c(x)))
a(d(x)) → d(a(x))
b(c(a(x))) → a(b(c(x)))
(20) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
c(a(x)) → a(c(x))
c(b(x)) → d(x)
c(R(x)) → c(b(R(x)))
a(d(x)) → d(a(x))
d(x) → b(a(x))
b(a(a(x))) → a(b(c(x)))
b(c(a(x))) → a(b(c(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