Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/2-matchbox)

The rewrite relation of the following TRS is considered.

c(c(x1)) a(a(a(b(b(b(x1)))))) (1)
b(b(b(a(x1)))) b(b(b(b(b(b(b(b(x1)))))))) (2)
b(b(c(c(x1)))) c(c(c(a(a(a(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 (2 x 2)-matrices with strict dimension 1 over the arctic semiring over the integers
[b(x1)] =
0 0
0 0
· x1 +
-∞ -∞
-∞ -∞
[c(x1)] =
0 1
0 0
· x1 +
-∞ -∞
-∞ -∞
[a(x1)] =
0 0
-∞ -∞
· x1 +
-∞ -∞
-∞ -∞
all of the following rules can be deleted.
c(c(x1)) a(a(a(b(b(b(x1)))))) (1)

1.1 Dependency Pair Transformation

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

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.