(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x) → n(c(c(x)))
c(f(x)) → f(c(c(x)))
c(c(x)) → c(x)
n(s(x)) → f(s(s(x)))
n(f(x)) → f(n(x))
Q is empty.
(1) QTRS Reverse (EQUIVALENT transformation)
We applied the QTRS Reverse Processor [REVERSE].
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x) → c(c(n(x)))
f(c(x)) → c(c(f(x)))
c(c(x)) → c(x)
s(n(x)) → s(s(f(x)))
f(n(x)) → n(f(x))
Q is empty.
(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:
F(x) → C(c(n(x)))
F(x) → C(n(x))
F(c(x)) → C(c(f(x)))
F(c(x)) → C(f(x))
F(c(x)) → F(x)
S(n(x)) → S(s(f(x)))
S(n(x)) → S(f(x))
S(n(x)) → F(x)
F(n(x)) → F(x)
The TRS R consists of the following rules:
f(x) → c(c(n(x)))
f(c(x)) → c(c(f(x)))
c(c(x)) → c(x)
s(n(x)) → s(s(f(x)))
f(n(x)) → n(f(x))
Q is empty.
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 5 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(n(x)) → F(x)
F(c(x)) → F(x)
The TRS R consists of the following rules:
f(x) → c(c(n(x)))
f(c(x)) → c(c(f(x)))
c(c(x)) → c(x)
s(n(x)) → s(s(f(x)))
f(n(x)) → n(f(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(n(x)) → F(x)
F(c(x)) → F(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) 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:
- F(n(x)) → F(x)
The graph contains the following edges 1 > 1
- F(c(x)) → F(x)
The graph contains the following edges 1 > 1
(11) YES
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(n(x)) → S(f(x))
S(n(x)) → S(s(f(x)))
The TRS R consists of the following rules:
f(x) → c(c(n(x)))
f(c(x)) → c(c(f(x)))
c(c(x)) → c(x)
s(n(x)) → s(s(f(x)))
f(n(x)) → n(f(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
S(n(x)) → S(f(x))
S(n(x)) → S(s(f(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( S(x1) ) = max{0, 2x1 - 1} |
| POL( s(x1) ) = max{0, -2} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
f(x) → c(c(n(x)))
f(c(x)) → c(c(f(x)))
f(n(x)) → n(f(x))
s(n(x)) → s(s(f(x)))
c(c(x)) → c(x)
(14) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(x) → c(c(n(x)))
f(c(x)) → c(c(f(x)))
c(c(x)) → c(x)
s(n(x)) → s(s(f(x)))
f(n(x)) → n(f(x))
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