Certification Problem

Input (TPDB TRS_Standard/Beerendonk_07/13)

The rewrite relation of the following TRS is considered.

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

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.