(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(y, 0), 0) → y
c(c(y)) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), y)
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 0
POL(a(x1, x2)) = x1 + x2
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(y)) → y
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(y, 0), 0) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), y)
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:
C(a(c(c(y)), x)) → A(c(c(c(a(x, 0)))), y)
C(a(c(c(y)), x)) → C(c(c(a(x, 0))))
C(a(c(c(y)), x)) → C(c(a(x, 0)))
C(a(c(c(y)), x)) → C(a(x, 0))
C(a(c(c(y)), x)) → A(x, 0)
The TRS R consists of the following rules:
a(a(y, 0), 0) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), y)
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 1 SCC with 2 less nodes.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(a(c(c(y)), x)) → C(c(a(x, 0)))
C(a(c(c(y)), x)) → C(c(c(a(x, 0))))
C(a(c(c(y)), x)) → C(a(x, 0))
The TRS R consists of the following rules:
a(a(y, 0), 0) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) 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(a(c(c(y)), x)) → C(c(a(x, 0)))
C(a(c(c(y)), x)) → C(a(x, 0))
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(C(x1)) = x1
POL(a(x1, x2)) = x1 + x2
POL(c(x1)) = 1 + x1
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(a(c(c(y)), x)) → C(c(c(a(x, 0))))
The TRS R consists of the following rules:
a(a(y, 0), 0) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 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(a(c(c(y)), x)) → C(c(c(a(x, 0))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(a(x1, x2)) = | 2A | + | 1A | · | x1 | + | 4A | · | x2 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
a(a(y, 0), 0) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), y)
(10) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a(a(y, 0), 0) → y
c(a(c(c(y)), x)) → a(c(c(c(a(x, 0)))), y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(12) YES