(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(s(x), x) → f(s(x), round(s(x)))
round(0) → 0
round(0) → s(0)
round(s(0)) → s(0)
round(s(s(x))) → s(s(round(x)))
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(s(x), x) → F(s(x), round(s(x)))
F(s(x), x) → ROUND(s(x))
ROUND(s(s(x))) → ROUND(x)
The TRS R consists of the following rules:
f(s(x), x) → f(s(x), round(s(x)))
round(0) → 0
round(0) → s(0)
round(s(0)) → s(0)
round(s(s(x))) → s(s(round(x)))
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 1 less node.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ROUND(s(s(x))) → ROUND(x)
The TRS R consists of the following rules:
f(s(x), x) → f(s(x), round(s(x)))
round(0) → 0
round(0) → s(0)
round(s(0)) → s(0)
round(s(s(x))) → s(s(round(x)))
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:
ROUND(s(s(x))) → ROUND(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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:
- ROUND(s(s(x))) → ROUND(x)
The graph contains the following edges 1 > 1
(9) YES
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(x), x) → F(s(x), round(s(x)))
The TRS R consists of the following rules:
f(s(x), x) → f(s(x), round(s(x)))
round(0) → 0
round(0) → s(0)
round(s(0)) → s(0)
round(s(s(x))) → s(s(round(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) 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.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(s(x), x) → F(s(x), round(s(x)))
The TRS R consists of the following rules:
round(s(0)) → s(0)
round(s(s(x))) → s(s(round(x)))
round(0) → 0
round(0) → s(0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) NonMonReductionPairProof (EQUIVALENT transformation)
Using the following max-polynomial ordering, we can orient the general usable rules and all rules from P weakly and some rules from P strictly:
Polynomial interpretation with max [POLO,NEGPOLO,MAXPOLO]:
POL(0) = 1
POL(F(x1, x2)) = max(0, 1 + x1 - x2)
POL(round(x1)) = x1
POL(s(x1)) = 1 + x1
The following pairs can be oriented strictly and are deleted.
F(s(x), x) → F(s(x), round(s(x)))
The remaining pairs can at least be oriented weakly.
none
The following rules are usable:
s(0) → round(s(0))
s(s(round(x))) → round(s(s(x)))
0 → round(0)
s(0) → round(0)
(14) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
round(s(0)) → s(0)
round(s(s(x))) → s(s(round(x)))
round(0) → 0
round(0) → s(0)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(16) YES