Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_AG01/#4.30)

The rewrite relation of the following TRS is considered.

minus(x,0) x (1)
minus(s(x),s(y)) minus(x,y) (2)
le(0,y) true (3)
le(s(x),0) false (4)
le(s(x),s(y)) le(x,y) (5)
quot(x,s(y)) if_quot(le(s(y),x),x,s(y)) (6)
if_quot(true,x,y) s(quot(minus(x,y),y)) (7)
if_quot(false,x,y) 0 (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.
minus#(s(x),s(y)) minus#(x,y) (9)
le#(s(x),s(y)) le#(x,y) (10)
quot#(x,s(y)) if_quot#(le(s(y),x),x,s(y)) (11)
quot#(x,s(y)) le#(s(y),x) (12)
if_quot#(true,x,y) quot#(minus(x,y),y) (13)
if_quot#(true,x,y) minus#(x,y) (14)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.