Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-115)

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

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

and S is the following TRS.

b(c(a(x1))) → a(c(b(x1))) (3)
a(b(a(x1))) → c(a(a(x1))) (4)
b(c(b(x1))) → b(c(b(x1))) (5)
c(b(a(x1))) → c(a(a(x1))) (6)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
c#(c(c(x1))) → b#(c(c(x1))) (7)
c#(b(a(x1))) → c#(a(a(x1))) (8)
c#(b(a(x1))) → a#(a(x1)) (9)
c#(a(c(x1))) → c#(c(b(x1))) (10)
c#(a(c(x1))) → c#(b(x1)) (11)
c#(a(c(x1))) → b#(x1) (12)
b#(c(a(x1))) → c#(b(x1)) (13)
b#(c(a(x1))) → b#(x1) (14)
b#(c(a(x1))) → a#(c(b(x1))) (15)
a#(b(a(x1))) → c#(a(a(x1))) (16)
a#(b(a(x1))) → a#(a(x1)) (17)

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 +
1
[a(x1)] = x1 +
1
[c#(x1)] = x1 +
0
[b#(x1)] = x1 +
0
[a#(x1)] = x1 +
0
together with the usable rules
c(a(c(x1))) → c(c(b(x1))) (1)
c(c(c(x1))) → b(c(c(x1))) (2)
b(c(a(x1))) → a(c(b(x1))) (3)
a(b(a(x1))) → c(a(a(x1))) (4)
c(b(a(x1))) → c(a(a(x1))) (6)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
c#(b(a(x1))) → a#(a(x1)) (9)
c#(a(c(x1))) → c#(b(x1)) (11)
c#(a(c(x1))) → b#(x1) (12)
b#(c(a(x1))) → c#(b(x1)) (13)
b#(c(a(x1))) → b#(x1) (14)
a#(b(a(x1))) → a#(a(x1)) (17)
and no rules could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.