Certification Problem

Input (TPDB SRS_Standard/Gebhardt_06/04)

The rewrite relation of the following TRS is considered.

0(0(0(0(x1)))) → 0(1(0(1(x1)))) (1)
0(1(0(1(x1)))) → 0(0(1(0(x1)))) (2)

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
0(0(0(0(x1)))) → 1(0(1(0(x1)))) (3)
1(0(1(0(x1)))) → 0(1(0(0(x1)))) (4)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
0#(0(0(0(x1)))) → 1#(0(1(0(x1)))) (5)
0#(0(0(0(x1)))) → 0#(1(0(x1))) (6)
0#(0(0(0(x1)))) → 1#(0(x1)) (7)
1#(0(1(0(x1)))) → 0#(1(0(0(x1)))) (8)
1#(0(1(0(x1)))) → 1#(0(0(x1))) (9)
1#(0(1(0(x1)))) → 0#(0(x1)) (10)

1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[0#(x1)] = 1 + 1 · x1
[0(x1)] = 1 + 1 · x1
[1#(x1)] = 1 + 1 · x1
[1(x1)] = 1 + 1 · x1
the pairs
0#(0(0(0(x1)))) → 0#(1(0(x1))) (6)
0#(0(0(0(x1)))) → 1#(0(x1)) (7)
1#(0(1(0(x1)))) → 1#(0(0(x1))) (9)
1#(0(1(0(x1)))) → 0#(0(x1)) (10)
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
[0#(x1)] =
-∞
-∞
-∞
+
0 0 0
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[0(x1)] =
0
0
1
+
-∞ 0 -∞
-∞ -∞ 0
0 -∞ 1
· x1
[1#(x1)] =
-∞
-∞
-∞
+
1 0 -∞
-∞ -∞ -∞
-∞ -∞ -∞
· x1
[1(x1)] =
0
0
0
+
1 0 -∞
-∞ 0 0
0 -∞ -∞
· x1
the pair
0#(0(0(0(x1)))) → 1#(0(1(0(x1)))) (5)
could be deleted.

1.1.1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals
[1#(x1)] = 1 + 1 · x1
[0(x1)] = 1 + 1 · x1
[1(x1)] = 1 + 1 · x1
[0#(x1)] = 0
having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
1#(0(1(0(x1)))) → 0#(1(0(0(x1)))) (8)
could be deleted.

1.1.1.1.1.1 P is empty

There are no pairs anymore.