(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
r(r(x)) → s(r(x))
r(s(x)) → s(r(x))
r(n(x)) → s(r(x))
r(b(x)) → u(s(b(x)))
r(u(x)) → u(r(x))
s(u(x)) → u(s(x))
n(u(x)) → u(n(x))
t(r(u(x))) → t(c(r(x)))
t(s(u(x))) → t(c(r(x)))
t(n(u(x))) → t(c(r(x)))
c(u(x)) → u(c(x))
c(s(x)) → s(c(x))
c(r(x)) → r(c(x))
c(n(x)) → n(c(x))
c(n(x)) → 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:
r(r(x)) → r(s(x))
s(r(x)) → r(s(x))
n(r(x)) → r(s(x))
b(r(x)) → b(s(u(x)))
u(r(x)) → r(u(x))
u(s(x)) → s(u(x))
u(n(x)) → n(u(x))
u(r(t(x))) → r(c(t(x)))
u(s(t(x))) → r(c(t(x)))
u(n(t(x))) → r(c(t(x)))
u(c(x)) → c(u(x))
s(c(x)) → c(s(x))
r(c(x)) → c(r(x))
n(c(x)) → c(n(x))
n(c(x)) → n(x)
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(b(x1)) = x1
POL(c(x1)) = x1
POL(n(x1)) = 1 + x1
POL(r(x1)) = 1 + x1
POL(s(x1)) = x1
POL(t(x1)) = x1
POL(u(x1)) = 1 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
r(r(x)) → r(s(x))
n(r(x)) → r(s(x))
u(r(t(x))) → r(c(t(x)))
u(n(t(x))) → r(c(t(x)))
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
s(r(x)) → r(s(x))
b(r(x)) → b(s(u(x)))
u(r(x)) → r(u(x))
u(s(x)) → s(u(x))
u(n(x)) → n(u(x))
u(s(t(x))) → r(c(t(x)))
u(c(x)) → c(u(x))
s(c(x)) → c(s(x))
r(c(x)) → c(r(x))
n(c(x)) → c(n(x))
n(c(x)) → n(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:
S(r(x)) → R(s(x))
S(r(x)) → S(x)
B(r(x)) → B(s(u(x)))
B(r(x)) → S(u(x))
B(r(x)) → U(x)
U(r(x)) → R(u(x))
U(r(x)) → U(x)
U(s(x)) → S(u(x))
U(s(x)) → U(x)
U(n(x)) → N(u(x))
U(n(x)) → U(x)
U(s(t(x))) → R(c(t(x)))
U(c(x)) → U(x)
S(c(x)) → S(x)
R(c(x)) → R(x)
N(c(x)) → N(x)
The TRS R consists of the following rules:
s(r(x)) → r(s(x))
b(r(x)) → b(s(u(x)))
u(r(x)) → r(u(x))
u(s(x)) → s(u(x))
u(n(x)) → n(u(x))
u(s(t(x))) → r(c(t(x)))
u(c(x)) → c(u(x))
s(c(x)) → c(s(x))
r(c(x)) → c(r(x))
n(c(x)) → c(n(x))
n(c(x)) → n(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 5 SCCs with 7 less nodes.
(8) Complex Obligation (AND)
(9) 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:
s(r(x)) → r(s(x))
b(r(x)) → b(s(u(x)))
u(r(x)) → r(u(x))
u(s(x)) → s(u(x))
u(n(x)) → n(u(x))
u(s(t(x))) → r(c(t(x)))
u(c(x)) → c(u(x))
s(c(x)) → c(s(x))
r(c(x)) → c(r(x))
n(c(x)) → c(n(x))
n(c(x)) → n(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:
N(c(x)) → N(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:
- N(c(x)) → N(x)
The graph contains the following edges 1 > 1
(13) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
R(c(x)) → R(x)
The TRS R consists of the following rules:
s(r(x)) → r(s(x))
b(r(x)) → b(s(u(x)))
u(r(x)) → r(u(x))
u(s(x)) → s(u(x))
u(n(x)) → n(u(x))
u(s(t(x))) → r(c(t(x)))
u(c(x)) → c(u(x))
s(c(x)) → c(s(x))
r(c(x)) → c(r(x))
n(c(x)) → c(n(x))
n(c(x)) → n(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:
R(c(x)) → R(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:
- R(c(x)) → R(x)
The graph contains the following edges 1 > 1
(18) YES
(19) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(c(x)) → S(x)
S(r(x)) → S(x)
The TRS R consists of the following rules:
s(r(x)) → r(s(x))
b(r(x)) → b(s(u(x)))
u(r(x)) → r(u(x))
u(s(x)) → s(u(x))
u(n(x)) → n(u(x))
u(s(t(x))) → r(c(t(x)))
u(c(x)) → c(u(x))
s(c(x)) → c(s(x))
r(c(x)) → c(r(x))
n(c(x)) → c(n(x))
n(c(x)) → n(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:
S(c(x)) → S(x)
S(r(x)) → S(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:
- S(c(x)) → S(x)
The graph contains the following edges 1 > 1
- S(r(x)) → S(x)
The graph contains the following edges 1 > 1
(23) YES
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
U(s(x)) → U(x)
U(r(x)) → U(x)
U(n(x)) → U(x)
U(c(x)) → U(x)
The TRS R consists of the following rules:
s(r(x)) → r(s(x))
b(r(x)) → b(s(u(x)))
u(r(x)) → r(u(x))
u(s(x)) → s(u(x))
u(n(x)) → n(u(x))
u(s(t(x))) → r(c(t(x)))
u(c(x)) → c(u(x))
s(c(x)) → c(s(x))
r(c(x)) → c(r(x))
n(c(x)) → c(n(x))
n(c(x)) → n(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:
U(s(x)) → U(x)
U(r(x)) → U(x)
U(n(x)) → U(x)
U(c(x)) → U(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(27) 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:
- U(s(x)) → U(x)
The graph contains the following edges 1 > 1
- U(r(x)) → U(x)
The graph contains the following edges 1 > 1
- U(n(x)) → U(x)
The graph contains the following edges 1 > 1
- U(c(x)) → U(x)
The graph contains the following edges 1 > 1
(28) YES
(29) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(r(x)) → B(s(u(x)))
The TRS R consists of the following rules:
s(r(x)) → r(s(x))
b(r(x)) → b(s(u(x)))
u(r(x)) → r(u(x))
u(s(x)) → s(u(x))
u(n(x)) → n(u(x))
u(s(t(x))) → r(c(t(x)))
u(c(x)) → c(u(x))
s(c(x)) → c(s(x))
r(c(x)) → c(r(x))
n(c(x)) → c(n(x))
n(c(x)) → n(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(30) 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.
(31) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(r(x)) → B(s(u(x)))
The TRS R consists of the following rules:
u(r(x)) → r(u(x))
u(s(x)) → s(u(x))
u(n(x)) → n(u(x))
u(s(t(x))) → r(c(t(x)))
u(c(x)) → c(u(x))
s(r(x)) → r(s(x))
s(c(x)) → c(s(x))
r(c(x)) → c(r(x))
n(c(x)) → c(n(x))
n(c(x)) → n(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(32) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
B(r(x)) → B(s(u(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(B(x1)) = x1
POL(c(x1)) = 0
POL(n(x1)) = 1
POL(r(x1)) = 1 + x1
POL(s(x1)) = x1
POL(t(x1)) = 1 + x1
POL(u(x1)) = x1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
u(r(x)) → r(u(x))
u(s(x)) → s(u(x))
u(n(x)) → n(u(x))
u(s(t(x))) → r(c(t(x)))
u(c(x)) → c(u(x))
s(r(x)) → r(s(x))
s(c(x)) → c(s(x))
r(c(x)) → c(r(x))
n(c(x)) → c(n(x))
n(c(x)) → n(x)
(33) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
u(r(x)) → r(u(x))
u(s(x)) → s(u(x))
u(n(x)) → n(u(x))
u(s(t(x))) → r(c(t(x)))
u(c(x)) → c(u(x))
s(r(x)) → r(s(x))
s(c(x)) → c(s(x))
r(c(x)) → c(r(x))
n(c(x)) → c(n(x))
n(c(x)) → n(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(34) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(35) YES