Certification Problem

Input (TPDB TRS_Standard/GTSSK07/cade14)

The rewrite relation of the following TRS is considered.

diff(x,y) cond1(equal(x,y),x,y) (1)
cond1(true,x,y) 0 (2)
cond1(false,x,y) cond2(gt(x,y),x,y) (3)
cond2(true,x,y) s(diff(x,s(y))) (4)
cond2(false,x,y) s(diff(s(x),y)) (5)
gt(0,v) false (6)
gt(s(u),0) true (7)
gt(s(u),s(v)) gt(u,v) (8)
equal(0,0) true (9)
equal(s(x),0) false (10)
equal(0,s(y)) false (11)
equal(s(x),s(y)) equal(x,y) (12)

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.
diff#(x,y) cond1#(equal(x,y),x,y) (13)
diff#(x,y) equal#(x,y) (14)
cond1#(false,x,y) cond2#(gt(x,y),x,y) (15)
cond1#(false,x,y) gt#(x,y) (16)
cond2#(true,x,y) diff#(x,s(y)) (17)
cond2#(false,x,y) diff#(s(x),y) (18)
gt#(s(u),s(v)) gt#(u,v) (19)
equal#(s(x),s(y)) equal#(x,y) (20)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.