Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z077)

The rewrite relation of the following TRS is considered.

f(0(x1)) s(0(x1)) (1)
d(0(x1)) 0(x1) (2)
d(s(x1)) s(s(d(x1))) (3)
f(s(x1)) d(f(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
[f(x1)] = 1 · x1 + 1
[0(x1)] = 1 · x1
[s(x1)] = 1 · x1
[d(x1)] = 1 · x1
all of the following rules can be deleted.
f(0(x1)) s(0(x1)) (1)

1.1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

10

Hence, it suffices to show innermost termination in the following.

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
d#(s(x1)) d#(x1) (5)
f#(s(x1)) d#(f(x1)) (6)
f#(s(x1)) f#(x1) (7)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.