(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
h(x, c(y, z)) → h(c(s(y), x), z)
h(c(s(x), c(s(0), y)), z) → h(y, c(s(0), c(x, z)))
Q is empty.
(1) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
h(x, c(y, z)) → h(c(s(y), x), z)
h(c(s(x), c(s(0), y)), z) → h(y, c(s(0), c(x, z)))
The signature Sigma is {
h}
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
h(x, c(y, z)) → h(c(s(y), x), z)
h(c(s(x), c(s(0), y)), z) → h(y, c(s(0), c(x, z)))
The set Q consists of the following terms:
h(x0, c(x1, x2))
h(c(s(x0), c(s(0), x1)), x2)
(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:
H(x, c(y, z)) → H(c(s(y), x), z)
H(c(s(x), c(s(0), y)), z) → H(y, c(s(0), c(x, z)))
The TRS R consists of the following rules:
h(x, c(y, z)) → h(c(s(y), x), z)
h(c(s(x), c(s(0), y)), z) → h(y, c(s(0), c(x, z)))
The set Q consists of the following terms:
h(x0, c(x1, x2))
h(c(s(x0), c(s(0), x1)), x2)
We have to consider all minimal (P,Q,R)-chains.
(5) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
H(x, c(y, z)) → H(c(s(y), x), z)
H(c(s(x), c(s(0), y)), z) → H(y, c(s(0), c(x, z)))
R is empty.
The set Q consists of the following terms:
h(x0, c(x1, x2))
h(c(s(x0), c(s(0), x1)), x2)
We have to consider all minimal (P,Q,R)-chains.
(7) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
h(x0, c(x1, x2))
h(c(s(x0), c(s(0), x1)), x2)
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
H(x, c(y, z)) → H(c(s(y), x), z)
H(c(s(x), c(s(0), y)), z) → H(y, c(s(0), c(x, z)))
R is empty.
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.
H(c(s(x), c(s(0), y)), z) → H(y, c(s(0), c(x, z)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:
POL(0) = [1]
POL(H(x1, x2)) = [4]x1 + [2]x2
POL(c(x1, x2)) = x1 + x2
POL(s(x1)) = [1/2]x1
The value of delta used in the strict ordering is 1.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
H(x, c(y, z)) → H(c(s(y), x), z)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- H(x, c(y, z)) → H(c(s(y), x), z)
The graph contains the following edges 2 > 2
(12) YES