(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
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(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))
L1(f(s(s(y)), f(z, w))) → F(s(0), f(y, f(s(z), w)))
L1(f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
L1(f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
F(x, f(s(s(y)), nil)) → F(y, f(s(0), nil))
F(x, f(s(s(y)), nil)) → F(s(0), nil)
The TRS R consists of the following rules:
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
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 5 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
The TRS R consists of the following rules:
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
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:
F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
The TRS R consists of the following rules:
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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:
F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(F(x1, x2)) = x1 + x2
POL(f(x1, x2)) = 2·x1 + x2
POL(nil) = 0
POL(s(x1)) = 1 + x1
(9) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(11) YES
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))
The TRS R consists of the following rules:
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) 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.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))
The TRS R consists of the following rules:
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( f(x1, x2) ) = max{0, 2x1 + x2 - 2} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
(16) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(18) YES