Certification Problem

Input (TPDB SRS_Standard/Zantema_06/13)

The rewrite relation of the following TRS is considered.

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

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
[b(x1)] = x1 +
1
[a(x1)] = x1 +
1/3
all of the following rules can be deleted.
a(a(a(x1))) a(a(x1)) (3)
b(b(x1)) a(b(a(x1))) (4)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(a(a(a(b(x1))))) b#(a(a(a(x1)))) (5)
b#(a(a(a(b(x1))))) a#(x1) (6)
b#(a(a(a(b(x1))))) a#(b(a(a(a(x1))))) (7)
b#(a(a(a(b(x1))))) a#(a(x1)) (8)
b#(a(a(a(b(x1))))) a#(a(b(a(a(a(x1)))))) (9)
b#(a(a(a(b(x1))))) a#(a(a(x1))) (10)
b#(a(a(a(b(x1))))) a#(a(a(b(a(a(a(x1))))))) (11)
a#(a(a(b(b(x1))))) b#(b(b(x1))) (12)

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
[b(x1)] = x1 +
1
[a(x1)] = x1 +
1/3
[b#(x1)] = x1 +
2/3
[a#(x1)] = x1 +
0
together with the usable rules
a(a(a(b(b(x1))))) b(b(b(x1))) (1)
b(a(a(a(b(x1))))) a(a(a(b(a(a(a(x1))))))) (2)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
b#(a(a(a(b(x1))))) b#(a(a(a(x1)))) (5)
b#(a(a(a(b(x1))))) a#(x1) (6)
b#(a(a(a(b(x1))))) a#(b(a(a(a(x1))))) (7)
b#(a(a(a(b(x1))))) a#(a(x1)) (8)
b#(a(a(a(b(x1))))) a#(a(b(a(a(a(x1)))))) (9)
b#(a(a(a(b(x1))))) a#(a(a(x1))) (10)
and no rules could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.