(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a, f(b(b(z, y), a))) → z
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(z, z))
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:
C(c(z, x, a), a, y) → F(f(c(y, a, f(c(z, y, x)))))
C(c(z, x, a), a, y) → F(c(y, a, f(c(z, y, x))))
C(c(z, x, a), a, y) → C(y, a, f(c(z, y, x)))
C(c(z, x, a), a, y) → F(c(z, y, x))
C(c(z, x, a), a, y) → C(z, y, x)
F(f(c(a, y, z))) → B(y, b(z, z))
F(f(c(a, y, z))) → B(z, z)
The TRS R consists of the following rules:
b(a, f(b(b(z, y), a))) → z
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(z, z))
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 1 SCC with 5 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(c(z, x, a), a, y) → C(z, y, x)
C(c(z, x, a), a, y) → C(y, a, f(c(z, y, x)))
The TRS R consists of the following rules:
b(a, f(b(b(z, y), a))) → z
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(z, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
C(c(z, x, a), a, y) → C(z, y, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(C(x1, x2, x3)) = | 1A | + | | · | x1 | + | | · | x2 | + | | · | x3 |
| POL(c(x1, x2, x3)) = | | + | | / | 0A | -I | 3A | \ |
| | | -I | 2A | 1A | | |
| \ | -I | 0A | 0A | / |
| · | x1 | + | | / | 0A | -I | -I | \ |
| | | -I | 1A | 2A | | |
| \ | -I | 0A | 0A | / |
| · | x2 | + | | / | 0A | 3A | 3A | \ |
| | | -I | 1A | 0A | | |
| \ | -I | 0A | 0A | / |
| · | x3 |
| POL(f(x1)) = | | + | | / | 0A | 1A | 1A | \ |
| | | -I | -I | 0A | | |
| \ | -I | -I | 0A | / |
| · | x1 |
| POL(b(x1, x2)) = | | + | | / | 0A | -I | 0A | \ |
| | | -I | 0A | 0A | | |
| \ | -I | 0A | -I | / |
| · | x1 | + | | / | 0A | -I | 0A | \ |
| | | -I | 0A | -I | | |
| \ | -I | 0A | 0A | / |
| · | x2 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(z, z))
b(a, f(b(b(z, y), a))) → z
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(c(z, x, a), a, y) → C(y, a, f(c(z, y, x)))
The TRS R consists of the following rules:
b(a, f(b(b(z, y), a))) → z
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(z, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
C(c(z, x, a), a, y) → C(y, a, f(c(z, y, x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(C(x1, x2, x3)) = | -I | + | | · | x1 | + | | · | x2 | + | | · | x3 |
| POL(c(x1, x2, x3)) = | | + | | / | -I | -I | -I | \ |
| | | -I | 0A | 0A | | |
| \ | -I | -I | -I | / |
| · | x1 | + | | / | -I | -I | -I | \ |
| | | -I | 0A | 0A | | |
| \ | 0A | 0A | -I | / |
| · | x2 | + | | / | -I | -I | -I | \ |
| | | 0A | 0A | 0A | | |
| \ | -I | -I | 0A | / |
| · | x3 |
| POL(f(x1)) = | | + | | / | 0A | -I | 0A | \ |
| | | -I | 0A | -I | | |
| \ | 0A | -I | -I | / |
| · | x1 |
| POL(b(x1, x2)) = | | + | | / | 0A | 0A | -I | \ |
| | | -I | -I | 0A | | |
| \ | -I | -I | -I | / |
| · | x1 | + | | / | -I | -I | 0A | \ |
| | | 0A | -I | -I | | |
| \ | -I | -I | 0A | / |
| · | x2 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(z, z))
b(a, f(b(b(z, y), a))) → z
(8) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
b(a, f(b(b(z, y), a))) → z
c(c(z, x, a), a, y) → f(f(c(y, a, f(c(z, y, x)))))
f(f(c(a, y, z))) → b(y, b(z, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(10) YES