Certification Problem

Input (TPDB SRS_Standard/Secret_05_SRS/jambox4)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
c(c(x1)) → b(a(x1)) (5)
b(x1) → a(a(x1)) (2)
b(b(b(x1))) → b(c(a(x1))) (6)
a(c(a(x1))) → c(c(a(x1))) (7)

1.1 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)] =
2 -∞
0 -∞
· x1 +
-∞ -∞
-∞ -∞
[c(x1)] =
0 3
0 -∞
· x1 +
-∞ -∞
-∞ -∞
[a(x1)] =
0 -∞
0 -∞
· x1 +
-∞ -∞
-∞ -∞
all of the following rules can be deleted.
b(b(b(x1))) → b(c(a(x1))) (6)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
c#(c(x1)) → a#(x1) (8)
c#(c(x1)) → b#(a(x1)) (9)
b#(x1) → a#(x1) (10)
b#(x1) → a#(a(x1)) (11)
a#(c(a(x1))) → c#(c(a(x1))) (12)

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 arctic semiring over the integers
[a#(x1)] =
-∞ 0
-∞ -∞
· x1 +
0 -∞
-∞ -∞
[b(x1)] =
0 1
-∞ 0
· x1 +
1 -∞
0 -∞
[c#(x1)] =
0 0
-∞ -∞
· x1 +
1 -∞
-∞ -∞
[b#(x1)] =
-∞ 1
-∞ -∞
· x1 +
1 -∞
-∞ -∞
[c(x1)] =
0 1
0 0
· x1 +
-∞ -∞
0 -∞
[a(x1)] =
0 1
-∞ 0
· x1 +
1 -∞
0 -∞
together with the usable rules
c(c(x1)) → b(a(x1)) (5)
b(x1) → a(a(x1)) (2)
a(c(a(x1))) → c(c(a(x1))) (7)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
c#(c(x1)) → a#(x1) (8)
b#(x1) → a#(x1) (10)
b#(x1) → a#(a(x1)) (11)
could be deleted.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.