Certification Problem

Input (TPDB TRS_Standard/Various_04/19)

The rewrite relation of the following TRS is considered.

:(x,x) e (1)
:(x,e) x (2)
i(:(x,y)) :(y,x) (3)
:(:(x,y),z) :(x,:(z,i(y))) (4)
:(e,x) i(x) (5)
i(i(x)) x (6)
i(e) e (7)
:(x,:(y,i(x))) i(y) (8)
:(x,:(y,:(i(x),z))) :(i(z),y) (9)
:(i(x),:(y,x)) i(y) (10)
:(i(x),:(y,:(x,z))) :(i(z),y) (11)

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.
:#(x,:(y,:(i(x),z))) i#(z) (12)
:#(e,x) i#(x) (13)
:#(:(x,y),z) :#(x,:(z,i(y))) (14)
:#(:(x,y),z) i#(y) (15)
i#(:(x,y)) :#(y,x) (16)
:#(:(x,y),z) :#(z,i(y)) (17)
:#(i(x),:(y,x)) i#(y) (18)
:#(i(x),:(y,:(x,z))) :#(i(z),y) (19)
:#(x,:(y,:(i(x),z))) :#(i(z),y) (20)
:#(x,:(y,i(x))) i#(y) (21)
:#(i(x),:(y,:(x,z))) i#(z) (22)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.