Certification Problem

Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-485)

The rewrite relation of the following TRS is considered.

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

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
a(a(x1)) → c(c(b(a(x1)))) (4)
a(c(x1)) → x1 (5)
b(c(x1)) → a(a(x1)) (6)

1.1 Dependency Pair Transformation

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

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)] =
0
-∞
-∞
+
0 -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[a(x1)] =
1
0
0
+
1 -∞ 0
0 -∞ -∞
0 0 -∞
· x1
[b#(x1)] =
1
-∞
-∞
+
-∞ 0 -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[c(x1)] =
0
1
0
+
-∞ 0 0
1 -∞ 0
0 -∞ 0
· x1
[b(x1)] =
-∞
-∞
-∞
+
0 1 -∞
-∞ 0 -∞
-∞ 0 -∞
· x1
the pair
b#(c(x1)) → a#(x1) (9)
could be deleted.

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
[a#(x1)] =
-∞
-∞
-∞
+
0 -∞ -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[a(x1)] =
1
0
0
+
1 -∞ 0
0 -∞ -∞
0 0 0
· x1
[b#(x1)] =
0
-∞
-∞
+
-∞ 0 -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[c(x1)] =
0
1
0
+
-∞ 0 -∞
1 -∞ 0
0 -∞ -∞
· x1
[b(x1)] =
1
-∞
-∞
+
0 1 -∞
-∞ 0 -∞
0 0 0
· x1
the pair
a#(a(x1)) → b#(a(x1)) (7)
could be deleted.

1.1.1.1.1 Monotonic Reduction Pair Processor

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

1.1.1.1.1.1 P is empty

There are no pairs anymore.