Certification Problem

Input (TPDB TRS_Standard/AProVE_08/id_inc)

The rewrite relation of the following TRS is considered.

f(s(x)) f(id_inc(c(x,x))) (1)
f(c(s(x),y)) g(c(x,y)) (2)
g(c(s(x),y)) g(c(y,x)) (3)
g(c(x,s(y))) g(c(y,x)) (4)
g(c(x,x)) f(x) (5)
id_inc(c(x,y)) c(id_inc(x),id_inc(y)) (6)
id_inc(s(x)) s(id_inc(x)) (7)
id_inc(0) 0 (8)
id_inc(0) s(0) (9)

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.
f#(s(x)) f#(id_inc(c(x,x))) (10)
f#(s(x)) id_inc#(c(x,x)) (11)
f#(c(s(x),y)) g#(c(x,y)) (12)
g#(c(s(x),y)) g#(c(y,x)) (13)
g#(c(x,s(y))) g#(c(y,x)) (14)
g#(c(x,x)) f#(x) (15)
id_inc#(c(x,y)) id_inc#(x) (16)
id_inc#(c(x,y)) id_inc#(y) (17)
id_inc#(s(x)) id_inc#(x) (18)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.