(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(x)) → b(b(b(x)))
a(f(x)) → f(a(a(x)))
b(b(x)) → c(c(a(c(x))))
d(b(x)) → d(a(b(x)))
c(c(x)) → d(d(d(x)))
b(d(x)) → d(b(x))
c(d(d(x))) → f(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(f(x)) → b(b(b(x)))
f(a(x)) → a(a(f(x)))
b(b(x)) → c(a(c(c(x))))
b(d(x)) → b(a(d(x)))
c(c(x)) → d(d(d(x)))
d(b(x)) → b(d(x))
d(d(c(x))) → f(x)
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(a(x1)) = x1
POL(b(x1)) = 53 + x1
POL(c(x1)) = 35 + x1
POL(d(x1)) = 23 + x1
POL(f(x1)) = 80 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
f(f(x)) → b(b(b(x)))
b(b(x)) → c(a(c(c(x))))
c(c(x)) → d(d(d(x)))
d(d(c(x))) → f(x)
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a(x)) → a(a(f(x)))
b(d(x)) → b(a(d(x)))
d(b(x)) → b(d(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(a(x)) → F(x)
B(d(x)) → B(a(d(x)))
D(b(x)) → B(d(x))
D(b(x)) → D(x)
The TRS R consists of the following rules:
f(a(x)) → a(a(f(x)))
b(d(x)) → b(a(d(x)))
d(b(x)) → b(d(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 2 SCCs with 2 less nodes.
(8) Complex Obligation (AND)
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
D(b(x)) → D(x)
The TRS R consists of the following rules:
f(a(x)) → a(a(f(x)))
b(d(x)) → b(a(d(x)))
d(b(x)) → b(d(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:
D(b(x)) → D(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:
- D(b(x)) → D(x)
The graph contains the following edges 1 > 1
(13) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a(x)) → F(x)
The TRS R consists of the following rules:
f(a(x)) → a(a(f(x)))
b(d(x)) → b(a(d(x)))
d(b(x)) → b(d(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:
F(a(x)) → F(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:
- F(a(x)) → F(x)
The graph contains the following edges 1 > 1
(18) YES