Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_AG01/#4.30b)

The rewrite relation of the following TRS is considered.

le(0,y) true (1)
le(s(x),0) false (2)
le(s(x),s(y)) le(x,y) (3)
minus(x,0) x (4)
minus(s(x),s(y)) minus(x,y) (5)
mod(0,y) 0 (6)
mod(s(x),0) 0 (7)
mod(s(x),s(y)) if_mod(le(y,x),s(x),s(y)) (8)
if_mod(true,x,y) mod(minus(x,y),y) (9)
if_mod(false,s(x),s(y)) s(x) (10)

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.
le#(s(x),s(y)) le#(x,y) (11)
minus#(s(x),s(y)) minus#(x,y) (12)
mod#(s(x),s(y)) if_mod#(le(y,x),s(x),s(y)) (13)
mod#(s(x),s(y)) le#(y,x) (14)
if_mod#(true,x,y) mod#(minus(x,y),y) (15)
if_mod#(true,x,y) minus#(x,y) (16)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.