Certification Problem

Input (TPDB TRS_Standard/Beerendonk_07/23)

The rewrite relation of the following TRS is considered.

cond1(true,x,y,z) cond2(gr(y,z),x,y,z) (1)
cond2(true,x,y,z) cond2(gr(y,z),x,p(y),z) (2)
cond2(false,x,y,z) cond1(gr(x,z),p(x),y,z) (3)
gr(0,x) false (4)
gr(s(x),0) true (5)
gr(s(x),s(y)) gr(x,y) (6)
p(0) 0 (7)
p(s(x)) x (8)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
cond1#(true,x,y,z) gr#(y,z) (9)
cond1#(true,x,y,z) cond2#(gr(y,z),x,y,z) (10)
cond2#(true,x,y,z) p#(y) (11)
cond2#(true,x,y,z) gr#(y,z) (12)
cond2#(true,x,y,z) cond2#(gr(y,z),x,p(y),z) (13)
cond2#(false,x,y,z) p#(x) (14)
cond2#(false,x,y,z) gr#(x,z) (15)
cond2#(false,x,y,z) cond1#(gr(x,z),p(x),y,z) (16)
gr#(s(x),s(y)) gr#(x,y) (17)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.