(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
cond(true, x) → cond(odd(x), p(p(p(x))))
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x
Q is empty.
(1) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x
The TRS R 2 is
cond(true, x) → cond(odd(x), p(p(p(x))))
The signature Sigma is {
cond}
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
cond(true, x) → cond(odd(x), p(p(p(x))))
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
cond(true, x0)
odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))
(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:
COND(true, x) → COND(odd(x), p(p(p(x))))
COND(true, x) → ODD(x)
COND(true, x) → P(p(p(x)))
COND(true, x) → P(p(x))
COND(true, x) → P(x)
ODD(s(s(x))) → ODD(x)
The TRS R consists of the following rules:
cond(true, x) → cond(odd(x), p(p(p(x))))
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
cond(true, x0)
odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ODD(s(s(x))) → ODD(x)
The TRS R consists of the following rules:
cond(true, x) → cond(odd(x), p(p(p(x))))
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
cond(true, x0)
odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(8) 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.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ODD(s(s(x))) → ODD(x)
R is empty.
The set Q consists of the following terms:
cond(true, x0)
odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(10) 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].
cond(true, x0)
odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ODD(s(s(x))) → ODD(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) 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:
- ODD(s(s(x))) → ODD(x)
The graph contains the following edges 1 > 1
(13) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND(true, x) → COND(odd(x), p(p(p(x))))
The TRS R consists of the following rules:
cond(true, x) → cond(odd(x), p(p(p(x))))
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
cond(true, x0)
odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(15) 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.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND(true, x) → COND(odd(x), p(p(p(x))))
The TRS R consists of the following rules:
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
cond(true, x0)
odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(17) 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].
cond(true, x0)
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND(true, x) → COND(odd(x), p(p(p(x))))
The TRS R consists of the following rules:
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(19) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
COND(true, x) → COND(odd(x), p(p(p(x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:
POL(0) = 0
POL(COND(x1, x2)) = [4]x1 + [4]x2
POL(false) = [2]
POL(odd(x1)) = [2] + [1/2]x1
POL(p(x1)) = [1/2]x1
POL(s(x1)) = [4] + [4]x1
POL(true) = [4]
The value of delta used in the strict ordering is 8.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x
(20) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(21) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(22) YES