Certification Problem

Input (TPDB TRS_Standard/Der95/11)

The rewrite relation of the following TRS is considered.

D(t) 1 (1)
D(constant) 0 (2)
D(+(x,y)) +(D(x),D(y)) (3)
D(*(x,y)) +(*(y,D(x)),*(x,D(y))) (4)
D(-(x,y)) -(D(x),D(y)) (5)
D(minus(x)) minus(D(x)) (6)
D(div(x,y)) -(div(D(x),y),div(*(x,D(y)),pow(y,2))) (7)
D(ln(x)) div(D(x),x) (8)
D(pow(x,y)) +(*(*(y,pow(x,-(y,1))),D(x)),*(*(pow(x,y),ln(x)),D(y))) (9)

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.
D#(pow(x,y)) D#(x) (10)
D#(-(x,y)) D#(x) (11)
D#(+(x,y)) D#(y) (12)
D#(*(x,y)) D#(x) (13)
D#(*(x,y)) D#(y) (14)
D#(-(x,y)) D#(y) (15)
D#(ln(x)) D#(x) (16)
D#(div(x,y)) D#(y) (17)
D#(pow(x,y)) D#(y) (18)
D#(minus(x)) D#(x) (19)
D#(+(x,y)) D#(x) (20)
D#(div(x,y)) D#(x) (21)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.