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