Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/secr8)

The rewrite relation of the following TRS is considered.

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

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

1.1 Subterm Criterion Processor

We use the projection to multisets
π(a#) = { 1, 1 }
π(b#) = { 1, 1 }
π(c) = { 1, 1 }
π(a) = { 1 }
π(b) = { 1 }
to remove the pairs:
b#(c(x1)) → a#(x1) (9)
b#(c(x1)) → a#(a(x1)) (10)
a#(c(x1)) → b#(x1) (11)
a#(c(x1)) → b#(b(x1)) (12)

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 arctic semiring over the integers
[a#(x1)] =
0 0
-∞ -∞
· x1 +
1 -∞
-∞ -∞
[a(x1)] =
1 0
0 -∞
· x1 +
0 -∞
0 -∞
[b#(x1)] =
0 0
-∞ -∞
· x1 +
0 -∞
-∞ -∞
[b(x1)] =
-∞ 0
0 1
· x1 +
0 -∞
1 -∞
[c(x1)] =
-∞ -∞
-∞ -∞
· x1 +
0 -∞
-∞ -∞
together with the usable rules
b(b(b(x1))) → a(x1) (1)
a(a(x1)) → a(b(a(x1))) (2)
b(c(x1)) → c(a(a(x1))) (3)
a(c(x1)) → c(b(b(x1))) (4)
a(a(a(x1))) → b(a(a(x1))) (5)
(w.r.t. the implicit argument filter of the reduction pair), the pair
b#(b(b(x1))) → a#(x1) (6)
could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.