Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-10)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[b(x1)] = 1 · x1 + -∞
[a(x1)] = 0 · x1 + -∞
all of the following rules can be deleted.
a(b(a(a(x1)))) a(a(a(a(x1)))) (1)

1.1 Dependency Pair Transformation

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

1.1.1 Subterm Criterion Processor

We use the projection to multisets
π(b#) = { 1, 1 }
π(b) = { 1, 1 }
π(a) = { 1, 1 }
to remove the pairs:
b#(a(b(a(x1)))) b#(x1) (4)
b#(a(b(a(x1)))) b#(b(x1)) (5)
b#(a(a(b(x1)))) b#(a(x1)) (6)

1.1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals
[b(x1)] =
0 0
2 2
· x1 +
0 0
2 0
[a(x1)] =
0 1
1 0
· x1 +
0 0
0 0
[b#(x1)] =
0 2
0 0
· x1 +
0 0
0 0
together with the usable rules
b(a(b(a(x1)))) a(a(b(b(x1)))) (2)
b(a(a(b(x1)))) b(a(b(a(x1)))) (3)
(w.r.t. the implicit argument filter of the reduction pair), the pair
b#(a(a(b(x1)))) b#(a(b(a(x1)))) (7)
could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.