Certification Problem

Input (TPDB TRS_Standard/SK90/2.23)

The rewrite relation of the following TRS is considered.

fac(0) 1 (1)
fac(s(x)) *(s(x),fac(x)) (2)
floop(0,y) y (3)
floop(s(x),y) floop(x,*(s(x),y)) (4)
*(x,0) 0 (5)
*(x,s(y)) +(*(x,y),x) (6)
+(x,0) x (7)
+(x,s(y)) s(+(x,y)) (8)
1 s(0) (9)
fac(0) s(0) (10)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
fac#(s(x)) fac#(x) (11)
+#(x,s(y)) +#(x,y) (12)
floop#(s(x),y) floop#(x,*(s(x),y)) (13)
fac#(0) 1# (14)
*#(x,s(y)) +#(*(x,y),x) (15)
fac#(s(x)) *#(s(x),fac(x)) (16)
floop#(s(x),y) *#(s(x),y) (17)
*#(x,s(y)) *#(x,y) (18)

1.1 Dependency Graph Processor

The dependency pairs are split into 4 components.