Certification Problem

Input (TPDB TRS_Standard/Rubio_04/polo2)

The rewrite relation of the following TRS is considered.

dx(X) one (1)
dx(a) zero (2)
dx(plus(ALPHA,BETA)) plus(dx(ALPHA),dx(BETA)) (3)
dx(times(ALPHA,BETA)) plus(times(BETA,dx(ALPHA)),times(ALPHA,dx(BETA))) (4)
dx(minus(ALPHA,BETA)) minus(dx(ALPHA),dx(BETA)) (5)
dx(neg(ALPHA)) neg(dx(ALPHA)) (6)
dx(div(ALPHA,BETA)) minus(div(dx(ALPHA),BETA),times(ALPHA,div(dx(BETA),exp(BETA,two)))) (7)
dx(ln(ALPHA)) div(dx(ALPHA),ALPHA) (8)
dx(exp(ALPHA,BETA)) plus(times(BETA,times(exp(ALPHA,minus(BETA,one)),dx(ALPHA))),times(exp(ALPHA,BETA),times(ln(ALPHA),dx(BETA)))) (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.
dx#(exp(ALPHA,BETA)) dx#(ALPHA) (10)
dx#(minus(ALPHA,BETA)) dx#(ALPHA) (11)
dx#(plus(ALPHA,BETA)) dx#(BETA) (12)
dx#(times(ALPHA,BETA)) dx#(ALPHA) (13)
dx#(times(ALPHA,BETA)) dx#(BETA) (14)
dx#(minus(ALPHA,BETA)) dx#(BETA) (15)
dx#(ln(ALPHA)) dx#(ALPHA) (16)
dx#(div(ALPHA,BETA)) dx#(BETA) (17)
dx#(exp(ALPHA,BETA)) dx#(BETA) (18)
dx#(neg(ALPHA)) dx#(ALPHA) (19)
dx#(plus(ALPHA,BETA)) dx#(ALPHA) (20)
dx#(div(ALPHA,BETA)) dx#(ALPHA) (21)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.