Certification Problem

Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-502)

The rewrite relation of the following TRS is considered.

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

1.1 Dependency Pair Transformation

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

1.1.1 Subterm Criterion Processor

We use the projection to multisets
π(c#) = { 1 }
π(b#) = { 1 }
π(a#) = { 1 }
π(c) = { 1, 1 }
π(b) = { 1, 1 }
π(a) = { 1, 1 }
to remove the pairs:
a#(a(x1)) → b#(x1) (6)
b#(a(x1)) → c#(x1) (8)
c#(b(x1)) → a#(x1) (10)

1.1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the arctic semiring over the integers
[b#(x1)] =
0 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1 +
0 -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
[b(x1)] =
0 0 0
0 0 0
0 0 0
· x1 +
0 -∞ -∞
0 -∞ -∞
0 -∞ -∞
[a#(x1)] =
-∞ 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1 +
0 -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
[c#(x1)] =
0 0 1
-∞ -∞ -∞
-∞ -∞ -∞
· x1 +
0 -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
[a(x1)] =
0 1 1
0 0 0
-∞ 0 0
· x1 +
1 -∞ -∞
0 -∞ -∞
0 -∞ -∞
[c(x1)] =
0 0 1
0 0 1
-∞ 0 0
· x1 +
0 -∞ -∞
1 -∞ -∞
-∞ -∞ -∞
together with the usable rules
a(a(x1)) → b(b(x1)) (1)
b(a(x1)) → c(c(x1)) (4)
c(b(x1)) → a(a(x1)) (5)
(w.r.t. the implicit argument filter of the reduction pair), the pair
c#(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.