Certification Problem

Input (TPDB TRS_Standard/Various_04/22)

The rewrite relation of the following TRS is considered.

f(x,0) s(0) (1)
f(s(x),s(y)) s(f(x,y)) (2)
g(0,x) g(f(x,x),x) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

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 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
f#(s(x),s(y)) f#(x,y) (4)
g#(0,x) g#(f(x,x),x) (5)
g#(0,x) f#(x,x) (6)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.