Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-16)

The rewrite relation of the following TRS is considered.

b(b(c(x1))) → a(a(b(x1))) (1)
b(b(a(x1))) → c(b(b(x1))) (2)
a(c(c(x1))) → c(a(b(x1))) (3)
a(b(a(x1))) → b(c(c(x1))) (4)
b(a(a(x1))) → c(b(b(x1))) (5)

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.
b#(b(c(x1))) → b#(x1) (6)
b#(b(c(x1))) → a#(b(x1)) (7)
b#(b(c(x1))) → a#(a(b(x1))) (8)
b#(b(a(x1))) → b#(x1) (9)
b#(b(a(x1))) → b#(b(x1)) (10)
b#(a(a(x1))) → b#(x1) (11)
b#(a(a(x1))) → b#(b(x1)) (12)
a#(c(c(x1))) → b#(x1) (13)
a#(c(c(x1))) → a#(b(x1)) (14)
a#(b(a(x1))) → b#(c(c(x1))) (15)

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
[b#(x1)] = x1 +
0
[a#(x1)] = x1 +
0
together with the usable rules
b(b(c(x1))) → a(a(b(x1))) (1)
b(b(a(x1))) → c(b(b(x1))) (2)
a(c(c(x1))) → c(a(b(x1))) (3)
a(b(a(x1))) → b(c(c(x1))) (4)
b(a(a(x1))) → c(b(b(x1))) (5)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
b#(b(c(x1))) → b#(x1) (6)
b#(b(c(x1))) → a#(b(x1)) (7)
b#(b(a(x1))) → b#(x1) (9)
b#(b(a(x1))) → b#(b(x1)) (10)
b#(a(a(x1))) → b#(x1) (11)
b#(a(a(x1))) → b#(b(x1)) (12)
a#(c(c(x1))) → b#(x1) (13)
a#(c(c(x1))) → a#(b(x1)) (14)
and no rules could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.