Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_CSR_05/Ex49_GM04)

The rewrite relation of the following TRS is considered.

minus(0,Y) 0 (1)
minus(s(X),s(Y)) minus(X,Y) (2)
geq(X,0) true (3)
geq(0,s(Y)) false (4)
geq(s(X),s(Y)) geq(X,Y) (5)
div(0,s(Y)) 0 (6)
div(s(X),s(Y)) if(geq(X,Y),s(div(minus(X,Y),s(Y))),0) (7)
if(true,X,Y) X (8)
if(false,X,Y) Y (9)

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) (10)
geq#(s(X),s(Y)) geq#(X,Y) (11)
div#(s(X),s(Y)) if#(geq(X,Y),s(div(minus(X,Y),s(Y))),0) (12)
div#(s(X),s(Y)) geq#(X,Y) (13)
div#(s(X),s(Y)) div#(minus(X,Y),s(Y)) (14)
div#(s(X),s(Y)) minus#(X,Y) (15)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.