Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-212)

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

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

and S is the following TRS.

a(b(a(x1))) → a(b(a(x1))) (3)
c(c(a(x1))) → c(c(c(x1))) (4)
c(a(a(x1))) → b(a(b(x1))) (5)
c(a(c(x1))) → b(a(c(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(a(x1))) → c#(x1) (7)
c#(c(a(x1))) → c#(c(x1)) (8)
c#(c(a(x1))) → c#(c(c(x1))) (9)
c#(a(c(x1))) → b#(a(c(x1))) (10)
c#(a(a(x1))) → b#(x1) (11)
c#(a(a(x1))) → b#(a(b(x1))) (12)
b#(b(c(x1))) → b#(x1) (13)
b#(b(c(x1))) → b#(b(x1)) (14)
b#(b(c(x1))) → b#(b(b(x1))) (15)
b#(b(b(x1))) → c#(a(a(x1))) (16)

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
together with the usable rules
b(b(b(x1))) → c(a(a(x1))) (1)
b(b(c(x1))) → b(b(b(x1))) (2)
c(c(a(x1))) → c(c(c(x1))) (4)
c(a(a(x1))) → b(a(b(x1))) (5)
c(a(c(x1))) → b(a(c(x1))) (6)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
c#(c(a(x1))) → c#(x1) (7)
c#(c(a(x1))) → c#(c(x1)) (8)
c#(a(a(x1))) → b#(x1) (11)
b#(b(c(x1))) → b#(x1) (13)
b#(b(c(x1))) → b#(b(x1)) (14)
and no rules could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.