Certification Problem

Input (TPDB TRS_Standard/MNZ_10/0)

The rewrite relation of the following TRS is considered.

s(s(0)) f(0) (1)
f(0) 0 (2)
s(s(s(0))) f(s(0)) (3)
f(s(0)) s(0) (4)
s(s(s(s(s(s(s(s(0)))))))) f(s(s(0))) (5)
f(s(s(0))) s(s(s(s(s(s(0)))))) (6)
g(x) h(x,x) (7)
s(x) h(x,0) (8)
s(x) h(0,x) (9)
f(g(x)) g(g(f(x))) (10)
g(s(x)) s(s(g(x))) (11)
h(f(x),g(x)) f(s(x)) (12)

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.
s#(s(0)) f#(0) (13)
s#(s(s(0))) f#(s(0)) (14)
s#(s(s(s(s(s(s(s(0)))))))) f#(s(s(0))) (15)
f#(s(s(0))) s#(s(s(s(s(s(0)))))) (16)
f#(s(s(0))) s#(s(s(s(s(0))))) (17)
f#(s(s(0))) s#(s(s(s(0)))) (18)
f#(s(s(0))) s#(s(s(0))) (19)
g#(x) h#(x,x) (20)
s#(x) h#(x,0) (21)
s#(x) h#(0,x) (22)
f#(g(x)) g#(g(f(x))) (23)
f#(g(x)) g#(f(x)) (24)
f#(g(x)) f#(x) (25)
g#(s(x)) s#(s(g(x))) (26)
g#(s(x)) s#(g(x)) (27)
g#(s(x)) g#(x) (28)
h#(f(x),g(x)) f#(s(x)) (29)
h#(f(x),g(x)) s#(x) (30)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.