Certification Problem

Input (TPDB TRS_Standard/Various_04/21)

The rewrite relation of the following TRS is considered.

+(p1,p1) p2 (1)
+(p1,+(p2,p2)) p5 (2)
+(p5,p5) p10 (3)
+(+(x,y),z) +(x,+(y,z)) (4)
+(p1,+(p1,x)) +(p2,x) (5)
+(p1,+(p2,+(p2,x))) +(p5,x) (6)
+(p2,p1) +(p1,p2) (7)
+(p2,+(p1,x)) +(p1,+(p2,x)) (8)
+(p2,+(p2,p2)) +(p1,p5) (9)
+(p2,+(p2,+(p2,x))) +(p1,+(p5,x)) (10)
+(p5,p1) +(p1,p5) (11)
+(p5,+(p1,x)) +(p1,+(p5,x)) (12)
+(p5,p2) +(p2,p5) (13)
+(p5,+(p2,x)) +(p2,+(p5,x)) (14)
+(p5,+(p5,x)) +(p10,x) (15)
+(p10,p1) +(p1,p10) (16)
+(p10,+(p1,x)) +(p1,+(p10,x)) (17)
+(p10,p2) +(p2,p10) (18)
+(p10,+(p2,x)) +(p2,+(p10,x)) (19)
+(p10,p5) +(p5,p10) (20)
+(p10,+(p5,x)) +(p5,+(p10,x)) (21)

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.
+#(p5,p2) +#(p2,p5) (22)
+#(p5,+(p1,x)) +#(p5,x) (23)
+#(p1,+(p1,x)) +#(p2,x) (24)
+#(p10,p5) +#(p5,p10) (25)
+#(p2,+(p2,+(p2,x))) +#(p1,+(p5,x)) (26)
+#(+(x,y),z) +#(x,+(y,z)) (27)
+#(p5,+(p5,x)) +#(p10,x) (28)
+#(p2,+(p2,+(p2,x))) +#(p5,x) (29)
+#(p10,+(p1,x)) +#(p1,+(p10,x)) (30)
+#(+(x,y),z) +#(y,z) (31)
+#(p10,p1) +#(p1,p10) (32)
+#(p5,+(p2,x)) +#(p2,+(p5,x)) (33)
+#(p2,p1) +#(p1,p2) (34)
+#(p5,+(p1,x)) +#(p1,+(p5,x)) (35)
+#(p2,+(p1,x)) +#(p2,x) (36)
+#(p10,+(p2,x)) +#(p10,x) (37)
+#(p10,+(p5,x)) +#(p10,x) (38)
+#(p10,p2) +#(p2,p10) (39)
+#(p2,+(p1,x)) +#(p1,+(p2,x)) (40)
+#(p10,+(p2,x)) +#(p2,+(p10,x)) (41)
+#(p2,+(p2,p2)) +#(p1,p5) (42)
+#(p1,+(p2,+(p2,x))) +#(p5,x) (43)
+#(p5,+(p2,x)) +#(p5,x) (44)
+#(p10,+(p1,x)) +#(p10,x) (45)
+#(p5,p1) +#(p1,p5) (46)
+#(p10,+(p5,x)) +#(p5,+(p10,x)) (47)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.