Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z087)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

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

1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a#(x1)] = 1 + 1 · x1
[a(x1)] = 1 + 1 · x1
[c#(x1)] = 1 + 1 · x1
[b(x1)] = 1 + 1 · x1
[b#(x1)] = 1 + 1 · x1
[c(x1)] = 1 + 1 · x1
the pairs
a#(a(x1)) → b#(x1) (5)
b#(b(x1)) → c#(x1) (7)
c#(c(x1)) → a#(x1) (9)
could be deleted.

1.1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals
[a#(x1)] =
1
-∞
-∞
+
1 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[a(x1)] =
0
0
0
+
1 0 -∞
0 0 1
1 0 0
· x1
[c#(x1)] =
0
-∞
-∞
+
0 -∞ 1
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[b(x1)] =
0
1
-∞
+
0 0 -∞
1 1 1
1 0 0
· x1
[b#(x1)] =
0
-∞
-∞
+
0 0 -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[c(x1)] =
0
-∞
0
+
-∞ 0 0
-∞ 0 1
1 0 1
· x1
the pair
c#(c(x1)) → b#(a(x1)) (8)
could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.