Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/secr8)

The rewrite relation of the following TRS is considered.

b(b(b(x1))) → a(x1) (1)
a(a(x1)) → a(b(a(x1))) (2)
b(c(x1)) → c(a(a(x1))) (3)
a(c(x1)) → c(b(b(x1))) (4)
a(a(a(x1))) → b(a(a(x1))) (5)

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.
b#(b(b(x1))) → a#(x1) (6)
a#(a(x1)) → a#(b(a(x1))) (7)
a#(a(x1)) → b#(a(x1)) (8)
b#(c(x1)) → a#(a(x1)) (9)
b#(c(x1)) → a#(x1) (10)
a#(c(x1)) → b#(b(x1)) (11)
a#(c(x1)) → b#(x1) (12)
a#(a(a(x1))) → b#(a(a(x1))) (13)

1.1 Monotonic Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[b(x1)] = 1 · x1
[a(x1)] = 1 · x1
[c(x1)] = 2 + 1 · x1
[b#(x1)] = 1 · x1
[a#(x1)] = 1 · x1
the pairs
b#(c(x1)) → a#(a(x1)) (9)
b#(c(x1)) → a#(x1) (10)
a#(c(x1)) → b#(b(x1)) (11)
a#(c(x1)) → b#(x1) (12)
and no rules 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
[b#(x1)] =
1
-∞
-∞
+
0 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[b(x1)] =
-∞
-∞
0
+
0 0 0
1 0 -∞
0 0 0
· x1
[a#(x1)] =
0
-∞
-∞
+
-∞ 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[a(x1)] =
0
1
0
+
-∞ 0 -∞
-∞ 1 0
0 0 0
· x1
[c(x1)] =
0
1
-∞
+
-∞ -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
the pair
b#(b(b(x1))) → a#(x1) (6)
could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.