Certification Problem

Input (TPDB TRS_Innermost/Mixed_innermost/test77)

The rewrite relation of the following TRS is considered.

+(X,0) X (1)
+(X,s(Y)) s(+(X,Y)) (2)
double(X) +(X,X) (3)
f(0,s(0),X) f(X,double(X),X) (4)
g(X,Y) X (5)
g(X,Y) Y (6)
The evaluation strategy is innermost.

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
+#(X,s(Y)) +#(X,Y) (7)
double#(X) +#(X,X) (8)
f#(0,s(0),X) f#(X,double(X),X) (9)
f#(0,s(0),X) double#(X) (10)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.