Certification Problem

Input (TPDB TRS_Standard/SK90/4.14)

The rewrite relation of the following TRS is considered.

p(s(x)) x (1)
s(p(x)) x (2)
+(0,y) y (3)
+(s(x),y) s(+(x,y)) (4)
+(p(x),y) p(+(x,y)) (5)
minus(0) 0 (6)
minus(s(x)) p(minus(x)) (7)
minus(p(x)) s(minus(x)) (8)
*(0,y) 0 (9)
*(s(x),y) +(*(x,y),y) (10)
*(p(x),y) +(*(x,y),minus(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.
*#(p(x),y) *#(x,y) (12)
*#(s(x),y) +#(*(x,y),y) (13)
+#(s(x),y) +#(x,y) (14)
+#(p(x),y) +#(x,y) (15)
minus#(p(x)) minus#(x) (16)
+#(s(x),y) s#(+(x,y)) (17)
*#(s(x),y) *#(x,y) (18)
minus#(p(x)) s#(minus(x)) (19)
minus#(s(x)) minus#(x) (20)
*#(p(x),y) minus#(y) (21)
*#(p(x),y) +#(*(x,y),minus(y)) (22)
+#(p(x),y) p#(+(x,y)) (23)
minus#(s(x)) p#(minus(x)) (24)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.