Certification Problem

Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-476)

The rewrite relation of the following TRS is considered.

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

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
a(a(x1)) → c(c(a(b(a(x1))))) (4)
a(c(x1)) → x1 (5)
b(c(x1)) → a(x1) (6)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(a(x1)) → b#(a(x1)) (7)
a#(a(x1)) → a#(b(a(x1))) (8)
b#(c(x1)) → a#(x1) (9)

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
[b#(x1)] =
0 -∞
-∞ -∞
· x1 +
0 -∞
-∞ -∞
[c(x1)] =
-∞ 0
0 -∞
· x1 +
1 -∞
1 -∞
[a#(x1)] =
-∞ 0
-∞ -∞
· x1 +
0 -∞
-∞ -∞
[a(x1)] =
-∞ 0
0 1
· x1 +
0 -∞
1 -∞
[b(x1)] =
0 -∞
1 0
· x1 +
0 -∞
0 -∞
together with the usable rules
a(a(x1)) → c(c(a(b(a(x1))))) (4)
a(c(x1)) → x1 (5)
b(c(x1)) → a(x1) (6)
(w.r.t. the implicit argument filter of the reduction pair), the pair
a#(a(x1)) → b#(a(x1)) (7)
could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.