Certification Problem

Input (TPDB SRS_Standard/Zantema_06/06)

The rewrite relation of the following TRS is considered.

b(a(b(x1))) → a(b(a(x1))) (1)
b(b(a(x1))) → b(b(b(x1))) (2)
c(a(x1)) → a(b(c(x1))) (3)
c(b(x1)) → 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
b(a(b(x1))) → a(b(a(x1))) (1)
a(b(b(x1))) → b(b(b(x1))) (5)
a(c(x1)) → c(b(a(x1))) (6)
b(c(x1)) → c(a(b(x1))) (7)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(a(b(x1))) → a#(b(a(x1))) (8)
b#(a(b(x1))) → b#(a(x1)) (9)
b#(a(b(x1))) → a#(x1) (10)
a#(b(b(x1))) → b#(b(b(x1))) (11)
a#(c(x1)) → b#(a(x1)) (12)
a#(c(x1)) → a#(x1) (13)
b#(c(x1)) → a#(b(x1)) (14)
b#(c(x1)) → b#(x1) (15)

1.1.1 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
a#(c(x1)) → b#(a(x1)) (12)
a#(c(x1)) → a#(x1) (13)
b#(c(x1)) → a#(b(x1)) (14)
b#(c(x1)) → b#(x1) (15)
and no rules could be deleted.

1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a#(x1)] = -2 + x1
[b#(x1)] = -2 + x1
[a(x1)] = 1 + 2 · x1
[b(x1)] = 1 + 2 · x1
[c(x1)] = -2
the pairs
b#(a(b(x1))) → b#(a(x1)) (9)
b#(a(b(x1))) → a#(x1) (10)
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
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[a(x1)] =
-∞
-∞
0
+
0 0 -∞
-∞ 0 -∞
0 1 0
· x1
[b(x1)] =
-∞
-∞
-∞
+
0 0 0
0 0 -∞
0 0 0
· x1
[a#(x1)] =
0
-∞
-∞
+
-∞ 1 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[c(x1)] =
0
-∞
0
+
-∞ -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
the pair
a#(b(b(x1))) → b#(b(b(x1))) (11)
could be deleted.

1.1.1.1.1.1 Monotonic Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[b(x1)] = 1 · x1
[a(x1)] = 1 · x1
[c(x1)] = 2 · x1
[b#(x1)] = 1 + 3 · x1
[a#(x1)] = 2 · x1
the pair
b#(a(b(x1))) → a#(b(a(x1))) (8)
and no rules could be deleted.

1.1.1.1.1.1.1 P is empty

There are no pairs anymore.