(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → b(c(x))
b(b(x)) → c(d(x))
c(c(x)) → d(d(d(x)))
d(c(x)) → b(f(x))
d(d(d(x))) → a(c(x))
f(f(x)) → f(b(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:
a(a(x)) → c(b(x))
b(b(x)) → d(c(x))
c(c(x)) → d(d(d(x)))
c(d(x)) → f(b(x))
d(d(d(x))) → c(a(x))
f(f(x)) → b(f(x))
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(a(x1)) = 21 + x1
POL(b(x1)) = 19 + x1
POL(c(x1)) = 23 + x1
POL(d(x1)) = 15 + x1
POL(f(x1)) = 19 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
c(c(x)) → d(d(d(x)))
d(d(d(x))) → c(a(x))
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → c(b(x))
b(b(x)) → d(c(x))
c(d(x)) → f(b(x))
f(f(x)) → b(f(x))
Q is empty.
(5) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(a(x1)) = 1 + x1
POL(b(x1)) = x1
POL(c(x1)) = x1
POL(d(x1)) = x1
POL(f(x1)) = x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
a(a(x)) → c(b(x))
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(x)) → d(c(x))
c(d(x)) → f(b(x))
f(f(x)) → b(f(x))
Q is empty.
(7) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(b(x)) → C(x)
C(d(x)) → F(b(x))
C(d(x)) → B(x)
F(f(x)) → B(f(x))
The TRS R consists of the following rules:
b(b(x)) → d(c(x))
c(d(x)) → f(b(x))
f(f(x)) → b(f(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
C(d(x)) → F(b(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(B(x1)) = 1
POL(C(x1)) = 1
POL(F(x1)) = x1
POL(b(x1)) = 0
POL(c(x1)) = x1
POL(d(x1)) = 0
POL(f(x1)) = 1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
b(b(x)) → d(c(x))
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(b(x)) → C(x)
C(d(x)) → B(x)
F(f(x)) → B(f(x))
The TRS R consists of the following rules:
b(b(x)) → d(c(x))
c(d(x)) → f(b(x))
f(f(x)) → b(f(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(d(x)) → B(x)
B(b(x)) → C(x)
The TRS R consists of the following rules:
b(b(x)) → d(c(x))
c(d(x)) → f(b(x))
f(f(x)) → b(f(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) 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.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
C(d(x)) → B(x)
B(b(x)) → C(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) 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:
- B(b(x)) → C(x)
The graph contains the following edges 1 > 1
- C(d(x)) → B(x)
The graph contains the following edges 1 > 1
(16) YES