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.