Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z114)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(c(x1)) b#(x1) (4)
a#(x1) b#(b(b(x1))) (5)
a#(x1) b#(b(x1)) (6)
a#(x1) b#(x1) (7)
b#(c(b(x1))) a#(c(x1)) (8)

1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a#(x1)] = 1 · x1
[c(x1)] = 1 + 1 · x1
[b#(x1)] = 1 · x1
[b(x1)] = 1 · x1
[a(x1)] = 1 · x1
the pair
a#(c(x1)) b#(x1) (4)
could be deleted.

1.1.1 Monotonic Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a(x1)] = 3 + 1 · x1
[c(x1)] = 3 · x1
[b(x1)] = 1 + 1 · x1
[a#(x1)] = 2 + 1 · x1
[b#(x1)] = 1 · x1
the pairs
a#(x1) b#(b(x1)) (6)
a#(x1) b#(x1) (7)
b#(c(b(x1))) a#(c(x1)) (8)
and the rule
b(c(b(x1))) a(c(x1)) (3)
could be deleted.

1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals
[b(x1)] = 1 · x1
[b#(x1)] = 1 · x1
[a#(x1)] = 1 · x1
having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

1.1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a#(x1)] = 1
[b#(x1)] = 0
[b(x1)] = 1 + 1 · x1
the pair
a#(x1) b#(b(b(x1))) (5)
could be deleted.

1.1.1.1.1.1 P is empty

There are no pairs anymore.