(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(f(a, b), c), x) → f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) → f(f(x, y), 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:
F(f(f(a, b), c), x) → F(b, f(a, f(c, f(b, x))))
F(f(f(a, b), c), x) → F(a, f(c, f(b, x)))
F(f(f(a, b), c), x) → F(c, f(b, x))
F(f(f(a, b), c), x) → F(b, x)
F(x, f(y, z)) → F(f(x, y), z)
F(x, f(y, z)) → F(x, y)
The TRS R consists of the following rules:
f(f(f(a, b), c), x) → f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) → f(f(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) 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(f(f(a, b), c), x) → F(c, f(b, x))
F(f(f(a, b), c), x) → F(b, x)
Used ordering: Polynomial interpretation [POLO]:
POL(F(x1, x2)) = x1 + x2
POL(a) = 1
POL(b) = 0
POL(c) = 0
POL(f(x1, x2)) = x1 + x2
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(f(f(a, b), c), x) → F(b, f(a, f(c, f(b, x))))
F(f(f(a, b), c), x) → F(a, f(c, f(b, x)))
F(x, f(y, z)) → F(f(x, y), z)
F(x, f(y, z)) → F(x, y)
The TRS R consists of the following rules:
f(f(f(a, b), c), x) → f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) → f(f(x, y), 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].
The following pairs can be oriented strictly and are deleted.
F(f(f(a, b), c), x) → F(b, f(a, f(c, f(b, x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(F(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(f(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
f(f(f(a, b), c), x) → f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) → f(f(x, y), z)
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(f(f(a, b), c), x) → F(a, f(c, f(b, x)))
F(x, f(y, z)) → F(f(x, y), z)
F(x, f(y, z)) → F(x, y)
The TRS R consists of the following rules:
f(f(f(a, b), c), x) → f(b, f(a, f(c, f(b, x))))
f(x, f(y, z)) → f(f(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) SemLabProof (SOUND transformation)
We found the following model for the rules of the TRS R.
Interpretation over the domain with elements from 0 to 1.c: 0
a: 1
f: 0
b: 0
F: 0
By semantic labelling [SEMLAB] we obtain the following labelled TRS.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F.0-0(f.0-0(f.1-0(a., b.), c.), x) → F.1-0(a., f.0-0(c., f.0-0(b., x)))
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.0-0(x, f.0-1(y, z)) → F.0-1(f.0-0(x, y), z)
F.0-0(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)
F.0-0(x, f.1-1(y, z)) → F.0-1(f.0-1(x, y), z)
F.1-0(x, f.0-0(y, z)) → F.0-0(f.1-0(x, y), z)
F.1-0(x, f.0-1(y, z)) → F.0-1(f.1-0(x, y), z)
F.1-0(x, f.1-0(y, z)) → F.0-0(f.1-1(x, y), z)
F.1-0(x, f.1-1(y, z)) → F.0-1(f.1-1(x, y), z)
F.0-1(f.0-0(f.1-0(a., b.), c.), x) → F.1-0(a., f.0-0(c., f.0-1(b., x)))
F.0-0(x, f.0-0(y, z)) → F.0-0(x, y)
F.0-0(x, f.0-1(y, z)) → F.0-0(x, y)
F.0-0(x, f.1-0(y, z)) → F.0-1(x, y)
F.0-0(x, f.1-1(y, z)) → F.0-1(x, y)
F.1-0(x, f.0-0(y, z)) → F.1-0(x, y)
F.1-0(x, f.0-1(y, z)) → F.1-0(x, y)
F.1-0(x, f.1-0(y, z)) → F.1-1(x, y)
F.1-0(x, f.1-1(y, z)) → F.1-1(x, y)
The TRS R consists of the following rules:
f.0-0(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.1-0(x, f.1-1(y, z)) → f.0-1(f.1-1(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
f.0-0(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.1-0(x, f.1-0(y, z)) → f.0-0(f.1-1(x, y), z)
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F.1-0(x, f.0-0(y, z)) → F.0-0(f.1-0(x, y), z)
F.0-0(f.0-0(f.1-0(a., b.), c.), x) → F.1-0(a., f.0-0(c., f.0-0(b., x)))
F.1-0(x, f.0-1(y, z)) → F.0-1(f.1-0(x, y), z)
F.0-1(f.0-0(f.1-0(a., b.), c.), x) → F.1-0(a., f.0-0(c., f.0-1(b., x)))
F.1-0(x, f.0-0(y, z)) → F.1-0(x, y)
F.1-0(x, f.1-0(y, z)) → F.0-0(f.1-1(x, y), z)
F.0-0(x, f.0-0(y, z)) → F.0-0(x, y)
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.0-0(x, f.0-1(y, z)) → F.0-0(x, y)
F.0-0(x, f.0-1(y, z)) → F.0-1(f.0-0(x, y), z)
F.0-0(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)
F.0-0(x, f.1-0(y, z)) → F.0-1(x, y)
F.0-0(x, f.1-1(y, z)) → F.0-1(x, y)
F.0-0(x, f.1-1(y, z)) → F.0-1(f.0-1(x, y), z)
F.1-0(x, f.0-1(y, z)) → F.1-0(x, y)
The TRS R consists of the following rules:
f.0-0(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.1-0(x, f.1-1(y, z)) → f.0-1(f.1-1(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
f.0-0(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.1-0(x, f.1-0(y, z)) → f.0-0(f.1-1(x, y), z)
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) 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.1-0(x, f.1-0(y, z)) → F.0-0(f.1-1(x, y), z)
F.0-0(x, f.1-0(y, z)) → F.0-0(f.0-1(x, y), z)
F.0-0(x, f.1-0(y, z)) → F.0-1(x, y)
F.0-0(x, f.1-1(y, z)) → F.0-1(x, y)
F.0-0(x, f.1-1(y, z)) → F.0-1(f.0-1(x, y), z)
Strictly oriented rules of the TRS R:
f.1-0(x, f.1-1(y, z)) → f.0-1(f.1-1(x, y), z)
f.0-0(x, f.1-0(y, z)) → f.0-0(f.0-1(x, y), z)
f.1-0(x, f.1-0(y, z)) → f.0-0(f.1-1(x, y), z)
f.0-0(x, f.1-1(y, z)) → f.0-1(f.0-1(x, y), z)
Used ordering: Polynomial interpretation [POLO]:
POL(F.0-0(x1, x2)) = x1 + x2
POL(F.0-1(x1, x2)) = x1 + x2
POL(F.1-0(x1, x2)) = 1 + x1 + x2
POL(a.) = 0
POL(b.) = 0
POL(c.) = 0
POL(f.0-0(x1, x2)) = x1 + x2
POL(f.0-1(x1, x2)) = x1 + x2
POL(f.1-0(x1, x2)) = 1 + x1 + x2
POL(f.1-1(x1, x2)) = 1 + x1 + x2
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F.1-0(x, f.0-0(y, z)) → F.0-0(f.1-0(x, y), z)
F.0-0(f.0-0(f.1-0(a., b.), c.), x) → F.1-0(a., f.0-0(c., f.0-0(b., x)))
F.1-0(x, f.0-1(y, z)) → F.0-1(f.1-0(x, y), z)
F.0-1(f.0-0(f.1-0(a., b.), c.), x) → F.1-0(a., f.0-0(c., f.0-1(b., x)))
F.1-0(x, f.0-0(y, z)) → F.1-0(x, y)
F.0-0(x, f.0-0(y, z)) → F.0-0(x, y)
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.0-0(x, f.0-1(y, z)) → F.0-0(x, y)
F.0-0(x, f.0-1(y, z)) → F.0-1(f.0-0(x, y), z)
F.1-0(x, f.0-1(y, z)) → F.1-0(x, y)
The TRS R consists of the following rules:
f.0-0(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) 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.0-0(x, f.0-1(y, z)) → F.0-0(x, y)
F.1-0(x, f.0-1(y, z)) → F.1-0(x, y)
Used ordering: Polynomial interpretation [POLO]:
POL(F.0-0(x1, x2)) = x1 + x2
POL(F.0-1(x1, x2)) = 1 + x1 + x2
POL(F.1-0(x1, x2)) = x1 + x2
POL(a.) = 0
POL(b.) = 0
POL(c.) = 0
POL(f.0-0(x1, x2)) = x1 + x2
POL(f.0-1(x1, x2)) = 1 + x1 + x2
POL(f.1-0(x1, x2)) = x1 + x2
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F.1-0(x, f.0-0(y, z)) → F.0-0(f.1-0(x, y), z)
F.0-0(f.0-0(f.1-0(a., b.), c.), x) → F.1-0(a., f.0-0(c., f.0-0(b., x)))
F.1-0(x, f.0-1(y, z)) → F.0-1(f.1-0(x, y), z)
F.0-1(f.0-0(f.1-0(a., b.), c.), x) → F.1-0(a., f.0-0(c., f.0-1(b., x)))
F.1-0(x, f.0-0(y, z)) → F.1-0(x, y)
F.0-0(x, f.0-0(y, z)) → F.0-0(x, y)
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.0-0(x, f.0-1(y, z)) → F.0-1(f.0-0(x, y), z)
The TRS R consists of the following rules:
f.0-0(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
F.0-0(f.0-0(f.1-0(a., b.), c.), x) → F.1-0(a., f.0-0(c., f.0-0(b., x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(F.0-0(x1, x2)) = x1
POL(F.0-1(x1, x2)) = 0
POL(F.1-0(x1, x2)) = x2
POL(a.) = 0
POL(b.) = 1
POL(c.) = 0
POL(f.0-0(x1, x2)) = x1
POL(f.0-1(x1, x2)) = x1
POL(f.1-0(x1, x2)) = x2
The following usable rules [FROCOS05] were oriented:
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F.1-0(x, f.0-0(y, z)) → F.0-0(f.1-0(x, y), z)
F.1-0(x, f.0-1(y, z)) → F.0-1(f.1-0(x, y), z)
F.0-1(f.0-0(f.1-0(a., b.), c.), x) → F.1-0(a., f.0-0(c., f.0-1(b., x)))
F.1-0(x, f.0-0(y, z)) → F.1-0(x, y)
F.0-0(x, f.0-0(y, z)) → F.0-0(x, y)
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.0-0(x, f.0-1(y, z)) → F.0-1(f.0-0(x, y), z)
The TRS R consists of the following rules:
f.0-0(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
F.0-1(f.0-0(f.1-0(a., b.), c.), x) → F.1-0(a., f.0-0(c., f.0-1(b., x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(F.0-0(x1, x2)) = x1
POL(F.0-1(x1, x2)) = x1
POL(F.1-0(x1, x2)) = x1 + x2
POL(a.) = 1
POL(b.) = 1
POL(c.) = 0
POL(f.0-0(x1, x2)) = x1
POL(f.0-1(x1, x2)) = x1
POL(f.1-0(x1, x2)) = x1 + x2
The following usable rules [FROCOS05] were oriented:
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F.1-0(x, f.0-0(y, z)) → F.0-0(f.1-0(x, y), z)
F.1-0(x, f.0-1(y, z)) → F.0-1(f.1-0(x, y), z)
F.1-0(x, f.0-0(y, z)) → F.1-0(x, y)
F.0-0(x, f.0-0(y, z)) → F.0-0(x, y)
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.0-0(x, f.0-1(y, z)) → F.0-1(f.0-0(x, y), z)
The TRS R consists of the following rules:
f.0-0(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(19) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.
(20) Complex Obligation (AND)
(21) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.0-0(x, f.0-0(y, z)) → F.0-0(x, y)
The TRS R consists of the following rules:
f.0-0(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(22) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
F.0-0(x, f.0-0(y, z)) → F.0-0(f.0-0(x, y), z)
F.0-0(x, f.0-0(y, z)) → F.0-0(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(F.0-0(x1, x2)) = x2
POL(a.) = 0
POL(b.) = 0
POL(c.) = 0
POL(f.0-0(x1, x2)) = 1 + x1 + x2
POL(f.0-1(x1, x2)) = 0
POL(f.1-0(x1, x2)) = 0
The following usable rules [FROCOS05] were oriented:
none
(23) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f.0-0(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(24) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(25) TRUE
(26) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F.1-0(x, f.0-0(y, z)) → F.1-0(x, y)
The TRS R consists of the following rules:
f.0-0(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-0(b., x))))
f.1-0(x, f.0-0(y, z)) → f.0-0(f.1-0(x, y), z)
f.0-0(x, f.0-1(y, z)) → f.0-1(f.0-0(x, y), z)
f.0-1(f.0-0(f.1-0(a., b.), c.), x) → f.0-0(b., f.1-0(a., f.0-0(c., f.0-1(b., x))))
f.1-0(x, f.0-1(y, z)) → f.0-1(f.1-0(x, y), z)
f.0-0(x, f.0-0(y, z)) → f.0-0(f.0-0(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(27) UsableRulesReductionPairsProof (EQUIVALENT transformation)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
F.1-0(x, f.0-0(y, z)) → F.1-0(x, y)
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(F.1-0(x1, x2)) = x1 + x2
POL(f.0-0(x1, x2)) = x1 + x2
(28) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(29) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(30) TRUE