(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
p(a(x0), p(x1, p(x2, x3))) → p(x1, p(x0, p(a(x3), x3)))
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:
P(a(x0), p(x1, p(x2, x3))) → P(x1, p(x0, p(a(x3), x3)))
P(a(x0), p(x1, p(x2, x3))) → P(x0, p(a(x3), x3))
P(a(x0), p(x1, p(x2, x3))) → P(a(x3), x3)
The TRS R consists of the following rules:
p(a(x0), p(x1, p(x2, x3))) → p(x1, p(x0, p(a(x3), x3)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
P(a(x0), p(x1, p(x2, x3))) → P(x0, p(a(x3), x3))
P(a(x0), p(x1, p(x2, x3))) → P(a(x3), x3)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(P(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(p(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
p(a(x0), p(x1, p(x2, x3))) → p(x1, p(x0, p(a(x3), x3)))
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P(a(x0), p(x1, p(x2, x3))) → P(x1, p(x0, p(a(x3), x3)))
The TRS R consists of the following rules:
p(a(x0), p(x1, p(x2, x3))) → p(x1, p(x0, p(a(x3), x3)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) ForwardInstantiation (EQUIVALENT transformation)
By forward instantiating [JAR06] the rule
P(
a(
x0),
p(
x1,
p(
x2,
x3))) →
P(
x1,
p(
x0,
p(
a(
x3),
x3))) we obtained the following new rules [LPAR04]:
P(a(x0), p(a(y_0), p(x2, x3))) → P(a(y_0), p(x0, p(a(x3), x3)))
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P(a(x0), p(a(y_0), p(x2, x3))) → P(a(y_0), p(x0, p(a(x3), x3)))
The TRS R consists of the following rules:
p(a(x0), p(x1, p(x2, x3))) → p(x1, p(x0, p(a(x3), x3)))
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.P: 0
a: 1
p: 0
By semantic labelling [SEMLAB] we obtain the following labelled TRS.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P.1-0(a.0(x0), p.1-0(a.0(y_0), p.0-0(x2, x3))) → P.1-0(a.0(y_0), p.0-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.0(y_0), p.0-1(x2, x3))) → P.1-0(a.0(y_0), p.0-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.0(y_0), p.1-0(x2, x3))) → P.1-0(a.0(y_0), p.0-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.0(y_0), p.1-1(x2, x3))) → P.1-0(a.0(y_0), p.0-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.1(y_0), p.0-0(x2, x3))) → P.1-0(a.1(y_0), p.0-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.1(y_0), p.0-1(x2, x3))) → P.1-0(a.1(y_0), p.0-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.1(y_0), p.1-0(x2, x3))) → P.1-0(a.1(y_0), p.0-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.0(x0), p.1-0(a.1(y_0), p.1-1(x2, x3))) → P.1-0(a.1(y_0), p.0-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.0(y_0), p.0-0(x2, x3))) → P.1-0(a.0(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.0(y_0), p.0-1(x2, x3))) → P.1-0(a.0(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.0(y_0), p.1-0(x2, x3))) → P.1-0(a.0(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.0(y_0), p.1-1(x2, x3))) → P.1-0(a.0(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.0-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.0-1(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-1(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
The TRS R consists of the following rules:
p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
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 2 SCCs with 13 less nodes.
(10) Complex Obligation (AND)
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-1(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
The TRS R consists of the following rules:
p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-1(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-1(a.1(x3), x3)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(P.1-0(x1, x2)) = x1 + x2
POL(a.1(x1)) = 1 + x1
POL(p.1-0(x1, x2)) = x1
POL(p.1-1(x1, x2)) = 0
The following usable rules [FROCOS05] were oriented:
none
(13) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(14) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(15) TRUE
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.0-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
The TRS R consists of the following rules:
p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
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.
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.0-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
P.1-0(a.1(x0), p.1-0(a.1(y_0), p.1-0(x2, x3))) → P.1-0(a.1(y_0), p.1-0(x0, p.1-0(a.0(x3), x3)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(P.1-0(x1, x2)) = x1 + x2
POL(a.0(x1)) = 0
POL(a.1(x1)) = 1 + x1
POL(p.0-0(x1, x2)) = x2
POL(p.0-1(x1, x2)) = 0
POL(p.1-0(x1, x2)) = x1 + x2
POL(p.1-1(x1, x2)) = 0
The following usable rules [FROCOS05] were oriented:
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
(18) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
p.1-0(a.1(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-0(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.0-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.0-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.0(x0), p.0-0(x1, p.1-1(x2, x3))) → p.0-0(x1, p.0-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-1(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-1(a.1(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.0-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
p.1-0(a.1(x0), p.1-0(x1, p.1-0(x2, x3))) → p.1-0(x1, p.1-0(x0, p.1-0(a.0(x3), x3)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(19) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(20) TRUE
(21) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
P(a(x0), p(x1, p(x2, x3))) → P(x0, p(a(x3), x3))
P(a(x0), p(x1, p(x2, x3))) → P(a(x3), x3)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(P(x1, x2)) = x2
POL(a(x1)) = 0
POL(p(x1, x2)) = 1 + x2
The following usable rules [FROCOS05] were oriented:
p(a(x0), p(x1, p(x2, x3))) → p(x1, p(x0, p(a(x3), x3)))
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P(a(x0), p(x1, p(x2, x3))) → P(x1, p(x0, p(a(x3), x3)))
The TRS R consists of the following rules:
p(a(x0), p(x1, p(x2, x3))) → p(x1, p(x0, p(a(x3), x3)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.