(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
t(f(x)) → t(c(n(x)))
n(f(x)) → f(n(x))
o(f(x)) → f(o(x))
n(s(x)) → f(s(x))
o(s(x)) → f(s(x))
c(f(x)) → f(c(x))
c(n(x)) → n(c(x))
c(o(x)) → o(c(x))
c(o(x)) → o(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(t(x)) → n(c(t(x)))
f(n(x)) → n(f(x))
f(o(x)) → o(f(x))
s(n(x)) → s(f(x))
s(o(x)) → s(f(x))
f(c(x)) → c(f(x))
n(c(x)) → c(n(x))
o(c(x)) → c(o(x))
o(c(x)) → o(x)
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(c(x1)) = x1
POL(f(x1)) = x1
POL(n(x1)) = x1
POL(o(x1)) = 1 + x1
POL(s(x1)) = x1
POL(t(x1)) = x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
s(o(x)) → s(f(x))
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(t(x)) → n(c(t(x)))
f(n(x)) → n(f(x))
f(o(x)) → o(f(x))
s(n(x)) → s(f(x))
f(c(x)) → c(f(x))
n(c(x)) → c(n(x))
o(c(x)) → c(o(x))
o(c(x)) → o(x)
Q is empty.
(5) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(t(x)) → N(c(t(x)))
F(n(x)) → N(f(x))
F(n(x)) → F(x)
F(o(x)) → O(f(x))
F(o(x)) → F(x)
S(n(x)) → S(f(x))
S(n(x)) → F(x)
F(c(x)) → F(x)
N(c(x)) → N(x)
O(c(x)) → O(x)
The TRS R consists of the following rules:
f(t(x)) → n(c(t(x)))
f(n(x)) → n(f(x))
f(o(x)) → o(f(x))
s(n(x)) → s(f(x))
f(c(x)) → c(f(x))
n(c(x)) → c(n(x))
o(c(x)) → c(o(x))
o(c(x)) → o(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.
(8) Complex Obligation (AND)
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
O(c(x)) → O(x)
The TRS R consists of the following rules:
f(t(x)) → n(c(t(x)))
f(n(x)) → n(f(x))
f(o(x)) → o(f(x))
s(n(x)) → s(f(x))
f(c(x)) → c(f(x))
n(c(x)) → c(n(x))
o(c(x)) → c(o(x))
o(c(x)) → o(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) 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.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
O(c(x)) → O(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:
- O(c(x)) → O(x)
The graph contains the following edges 1 > 1
(13) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
N(c(x)) → N(x)
The TRS R consists of the following rules:
f(t(x)) → n(c(t(x)))
f(n(x)) → n(f(x))
f(o(x)) → o(f(x))
s(n(x)) → s(f(x))
f(c(x)) → c(f(x))
n(c(x)) → c(n(x))
o(c(x)) → c(o(x))
o(c(x)) → o(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) 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.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
N(c(x)) → N(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) 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:
- N(c(x)) → N(x)
The graph contains the following edges 1 > 1
(18) YES
(19) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(o(x)) → F(x)
F(n(x)) → F(x)
F(c(x)) → F(x)
The TRS R consists of the following rules:
f(t(x)) → n(c(t(x)))
f(n(x)) → n(f(x))
f(o(x)) → o(f(x))
s(n(x)) → s(f(x))
f(c(x)) → c(f(x))
n(c(x)) → c(n(x))
o(c(x)) → c(o(x))
o(c(x)) → o(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(20) 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.
(21) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(o(x)) → F(x)
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.
(22) 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(o(x)) → F(x)
The graph contains the following edges 1 > 1
- 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
(23) YES
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(n(x)) → S(f(x))
The TRS R consists of the following rules:
f(t(x)) → n(c(t(x)))
f(n(x)) → n(f(x))
f(o(x)) → o(f(x))
s(n(x)) → s(f(x))
f(c(x)) → c(f(x))
n(c(x)) → c(n(x))
o(c(x)) → c(o(x))
o(c(x)) → o(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(25) 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.
(26) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(n(x)) → S(f(x))
The TRS R consists of the following rules:
f(t(x)) → n(c(t(x)))
f(n(x)) → n(f(x))
f(o(x)) → o(f(x))
f(c(x)) → c(f(x))
o(c(x)) → c(o(x))
o(c(x)) → o(x)
n(c(x)) → c(n(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(27) 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))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(S(x1)) = x1
POL(c(x1)) = 0
POL(f(x1)) = x1
POL(n(x1)) = 1 + x1
POL(o(x1)) = 1
POL(t(x1)) = 1 + x1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
f(t(x)) → n(c(t(x)))
f(n(x)) → n(f(x))
f(o(x)) → o(f(x))
f(c(x)) → c(f(x))
o(c(x)) → c(o(x))
o(c(x)) → o(x)
n(c(x)) → c(n(x))
(28) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(t(x)) → n(c(t(x)))
f(n(x)) → n(f(x))
f(o(x)) → o(f(x))
f(c(x)) → c(f(x))
o(c(x)) → c(o(x))
o(c(x)) → o(x)
n(c(x)) → c(n(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(29) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(30) YES