Certification Problem

Input (TPDB TRS_Standard/AG01/#3.36)

The rewrite relation of the following TRS is considered.

minus(x,0) x (1)
minus(s(x),s(y)) minus(x,y) (2)
f(0) s(0) (3)
f(s(x)) minus(s(x),g(f(x))) (4)
g(0) 0 (5)
g(s(x)) minus(s(x),f(g(x))) (6)

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.
minus#(s(x),s(y)) minus#(x,y) (7)
f#(s(x)) minus#(s(x),g(f(x))) (8)
f#(s(x)) g#(f(x)) (9)
f#(s(x)) f#(x) (10)
g#(s(x)) minus#(s(x),f(g(x))) (11)
g#(s(x)) f#(g(x)) (12)
g#(s(x)) g#(x) (13)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.