Certification Problem

Input (TPDB TRS_Standard/Beerendonk_07/22)

The rewrite relation of the following TRS is considered.

cond1(true,x,y) cond2(gr(y,0),x,y) (1)
cond2(true,x,y) cond2(gr(y,0),x,p(y)) (2)
cond2(false,x,y) cond1(gr(x,0),p(x),y) (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) gr#(y,0) (9)
cond1#(true,x,y) cond2#(gr(y,0),x,y) (10)
cond2#(true,x,y) p#(y) (11)
cond2#(true,x,y) gr#(y,0) (12)
cond2#(true,x,y) cond2#(gr(y,0),x,p(y)) (13)
cond2#(false,x,y) p#(x) (14)
cond2#(false,x,y) gr#(x,0) (15)
cond2#(false,x,y) cond1#(gr(x,0),p(x),y) (16)
gr#(s(x),s(y)) gr#(x,y) (17)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.