Certification Problem

Input (TPDB TRS_Standard/SK90/4.60)

The rewrite relation of the following TRS is considered.

msort(nil) nil (1)
msort(.(x,y)) .(min(x,y),msort(del(min(x,y),.(x,y)))) (2)
min(x,nil) x (3)
min(x,.(y,z)) if(<=(x,y),min(x,z),min(y,z)) (4)
del(x,nil) nil (5)
del(x,.(y,z)) if(=(x,y),z,.(y,del(x,z))) (6)

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.
msort#(.(x,y)) min#(x,y) (7)
msort#(.(x,y)) msort#(del(min(x,y),.(x,y))) (8)
msort#(.(x,y)) del#(min(x,y),.(x,y)) (9)
min#(x,.(y,z)) min#(x,z) (10)
min#(x,.(y,z)) min#(y,z) (11)
del#(x,.(y,z)) del#(x,z) (12)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.