Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-70)

The relative rewrite relation R/S is considered where R is the following TRS

b(a(a(x1))) → c(b(b(x1))) (1)
a(a(a(x1))) → a(c(c(x1))) (2)
a(c(b(x1))) → a(a(b(x1))) (3)
c(a(c(x1))) → c(c(c(x1))) (4)

and S is the following TRS.

c(a(a(x1))) → c(a(a(x1))) (5)
c(c(a(x1))) → c(b(b(x1))) (6)
a(a(b(x1))) → b(a(b(x1))) (7)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[c(x1)] = x1 +
1
[b(x1)] = x1 +
0
[a(x1)] = x1 +
1
all of the following rules can be deleted.
b(a(a(x1))) → c(b(b(x1))) (1)
c(a(a(x1))) → c(a(a(x1))) (5)
c(c(a(x1))) → c(b(b(x1))) (6)
a(a(b(x1))) → b(a(b(x1))) (7)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
c#(a(c(x1))) → c#(c(x1)) (8)
c#(a(c(x1))) → c#(c(c(x1))) (9)
a#(c(b(x1))) → a#(b(x1)) (10)
a#(c(b(x1))) → a#(a(b(x1))) (11)
a#(a(a(x1))) → c#(x1) (12)
a#(a(a(x1))) → c#(c(x1)) (13)
a#(a(a(x1))) → a#(c(c(x1))) (14)

1.1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[c(x1)] = x1 +
1
[b(x1)] = x1 +
0
[a(x1)] = x1 +
1
[c#(x1)] = x1 +
0
[a#(x1)] = x1 +
1
together with the usable rules
a(a(a(x1))) → a(c(c(x1))) (2)
a(c(b(x1))) → a(a(b(x1))) (3)
c(a(c(x1))) → c(c(c(x1))) (4)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
c#(a(c(x1))) → c#(c(x1)) (8)
a#(c(b(x1))) → a#(b(x1)) (10)
a#(a(a(x1))) → c#(x1) (12)
a#(a(a(x1))) → c#(c(x1)) (13)
and no rules could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.