Certification Problem

Input (TPDB SRS_Standard/Gebhardt_06/18)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
1#(0(0(1(x1)))) → 0#(x1) (3)
1#(0(0(1(x1)))) → 0#(0(x1)) (4)
1#(0(0(1(x1)))) → 0#(0(0(x1))) (5)
1#(0(0(1(x1)))) → 0#(0(0(0(x1)))) (6)
0#(0(0(0(x1)))) → 1#(x1) (7)
0#(0(0(0(x1)))) → 1#(1(x1)) (8)
0#(0(0(0(x1)))) → 1#(0(1(1(x1)))) (9)
0#(0(0(0(x1)))) → 0#(1(1(x1))) (10)

1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[1(x1)] = x1 +
1
[0(x1)] = x1 +
1
[1#(x1)] = x1 +
0
[0#(x1)] = x1 +
0
together with the usable rules
0(0(0(0(x1)))) → 1(0(1(1(x1)))) (1)
1(0(0(1(x1)))) → 0(0(0(0(x1)))) (2)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
1#(0(0(1(x1)))) → 0#(x1) (3)
1#(0(0(1(x1)))) → 0#(0(x1)) (4)
1#(0(0(1(x1)))) → 0#(0(0(x1))) (5)
0#(0(0(0(x1)))) → 1#(x1) (7)
0#(0(0(0(x1)))) → 1#(1(x1)) (8)
0#(0(0(0(x1)))) → 0#(1(1(x1))) (10)
and no rules could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.