Certification Problem

Input (TPDB SRS_Standard/Secret_07_SRS/x03)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
c(b(a(x1))) → b(x1) (5)
b(b(c(x1))) → a(x1) (6)
c(x1) → b(x1) (3)
a(a(x1)) → c(a(b(c(x1)))) (7)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c(x1)] = 1 · x1 + 1
[b(x1)] = 1 · x1 + 1
[a(x1)] = 1 · x1 + 3
all of the following rules can be deleted.
c(b(a(x1))) → b(x1) (5)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(b(c(x1))) → a#(x1) (8)
c#(x1) → b#(x1) (9)
a#(a(x1)) → c#(a(b(c(x1)))) (10)
a#(a(x1)) → a#(b(c(x1))) (11)
a#(a(x1)) → b#(c(x1)) (12)
a#(a(x1)) → c#(x1) (13)

1.1.1.1 Monotonic Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[b(x1)] = 1 + 1 · x1
[c(x1)] = 1 + 1 · x1
[a(x1)] = 3 + 1 · x1
[b#(x1)] = 1 · x1
[a#(x1)] = 2 + 1 · x1
[c#(x1)] = 1 · x1
the pairs
a#(a(x1)) → a#(b(c(x1))) (11)
a#(a(x1)) → b#(c(x1)) (12)
a#(a(x1)) → c#(x1) (13)
and no rules could be deleted.

1.1.1.1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals
[b#(x1)] =
0
-∞
-∞
+
0 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[b(x1)] =
0
-∞
-∞
+
-∞ -∞ 0
-∞ -∞ 0
-∞ 0 -∞
· x1
[c(x1)] =
0
0
-∞
+
-∞ -∞ 0
0 1 0
-∞ 0 -∞
· x1
[a#(x1)] =
0
-∞
-∞
+
0 1 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[c#(x1)] =
0
-∞
-∞
+
0 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[a(x1)] =
0
0
-∞
+
0 0 -∞
0 1 -∞
-∞ 0 -∞
· x1
the pair
a#(a(x1)) → c#(a(b(c(x1)))) (10)
could be deleted.

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.