(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(x, a), a) → f(f(f(a, f(a, a)), a), 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:
F(f(x, a), a) → F(f(f(a, f(a, a)), a), x)
F(f(x, a), a) → F(f(a, f(a, a)), a)
F(f(x, a), a) → F(a, f(a, a))
F(f(x, a), a) → F(a, a)
The TRS R consists of the following rules:
f(f(x, a), a) → f(f(f(a, f(a, a)), a), 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 1 SCC with 3 less nodes.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(f(x, a), a) → F(f(f(a, f(a, a)), a), x)
The TRS R consists of the following rules:
f(f(x, a), a) → f(f(f(a, f(a, a)), a), x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) 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.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(f(x, a), a) → F(f(f(a, f(a, a)), a), x)
R is empty.
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.
F(f(x, a), a) → F(f(f(a, f(a, a)), a), x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( F(x1, x2) ) = x1 + x2 + 1 |
| POL( f(x1, x2) ) = max{0, x1 - 1} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none
(8) Obligation:
Q DP problem:
P is empty.
R is empty.
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