Certification Problem

Input (TPDB SRS_Standard/Secret_07_SRS/dj)

The rewrite relation of the following TRS is considered.

1(0(x1)) 0(0(0(1(x1)))) (1)
0(1(x1)) 1(x1) (2)
1(1(x1)) 0(0(0(0(x1)))) (3)
0(0(x1)) 0(x1) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals
[1(x1)] = 1 · x1 + 1
[0(x1)] = 1 · x1
all of the following rules can be deleted.
1(1(x1)) 0(0(0(0(x1)))) (3)

1.1 Dependency Pair Transformation

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

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.