Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-373)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(b(a(a(x1)))) b#(x1) (4)
b#(b(a(a(x1)))) a#(b(x1)) (5)
b#(b(a(a(x1)))) a#(a(b(x1))) (6)
b#(b(a(a(x1)))) a#(a(a(b(x1)))) (7)
a#(a(a(b(x1)))) a#(x1) (8)
a#(a(a(b(x1)))) b#(a(x1)) (9)
a#(a(a(b(x1)))) a#(b(a(x1))) (10)
a#(a(a(b(x1)))) b#(a(b(a(x1)))) (11)
a#(b(a(b(x1)))) b#(b(x1)) (12)
a#(b(a(b(x1)))) a#(b(b(x1))) (13)
a#(b(a(b(x1)))) a#(a(b(b(x1)))) (14)

1.1 Subterm Criterion Processor

We use the projection to multisets
π(a#) = { 1, 1 }
π(b#) = { 1, 1 }
π(b) = { 1, 1 }
π(a) = { 1, 1 }
to remove the pairs:
b#(b(a(a(x1)))) b#(x1) (4)
b#(b(a(a(x1)))) a#(b(x1)) (5)
b#(b(a(a(x1)))) a#(a(b(x1))) (6)
a#(a(a(b(x1)))) a#(x1) (8)
a#(a(a(b(x1)))) b#(a(x1)) (9)
a#(a(a(b(x1)))) a#(b(a(x1))) (10)
a#(b(a(b(x1)))) b#(b(x1)) (12)
a#(b(a(b(x1)))) a#(b(b(x1))) (13)

1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[b(x1)] =
1 0 1
1 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[a#(x1)] =
0 1 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 0 1
1 0 0
0 1 0
· x1 +
1 0 0
0 0 0
0 0 0
[b#(x1)] =
0 1 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
together with the usable rules
b(b(a(a(x1)))) a(a(a(b(x1)))) (1)
a(a(a(b(x1)))) b(a(b(a(x1)))) (2)
a(b(a(b(x1)))) a(a(b(b(x1)))) (3)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
b#(b(a(a(x1)))) a#(a(a(b(x1)))) (7)
a#(b(a(b(x1)))) a#(a(b(b(x1)))) (14)
could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.