Certification Problem

Input (TPDB SRS_Standard/Gebhardt_06/09)

The rewrite relation of the following TRS is considered.

0(0(0(0(x1)))) 0(1(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 AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

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

1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[0#(x1)] = -2 + 2 · x1
[1#(x1)] = -2 + 2 · x1
[1(x1)] = 1 + x1
[0(x1)] = 1 + x1
the pairs
0#(0(0(0(x1)))) 1#(1(1(x1))) (4)
0#(0(0(0(x1)))) 1#(1(x1)) (5)
0#(0(0(0(x1)))) 1#(x1) (6)
1#(0(0(1(x1)))) 0#(0(0(x1))) (8)
1#(0(0(1(x1)))) 0#(0(x1)) (9)
1#(0(0(1(x1)))) 0#(x1) (10)
could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.