Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/tpa10)

The rewrite relation of the following TRS is considered.

min(0,y) 0 (1)
min(x,0) 0 (2)
min(s(x),s(y)) s(min(x,y)) (3)
max(0,y) y (4)
max(x,0) x (5)
max(s(x),s(y)) s(max(x,y)) (6)
+(0,y) y (7)
+(s(x),y) s(+(x,y)) (8)
-(x,0) x (9)
-(s(x),s(y)) -(x,y) (10)
*(x,0) 0 (11)
*(x,s(y)) +(x,*(x,y)) (12)
p(s(x)) x (13)
f(s(x),s(y)) f(-(min(s(x),s(y)),max(s(x),s(y))),*(s(x),s(y))) (14)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

10

Hence, it suffices to show innermost termination in the following.

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
min#(s(x),s(y)) min#(x,y) (15)
max#(s(x),s(y)) max#(x,y) (16)
+#(s(x),y) +#(x,y) (17)
-#(s(x),s(y)) -#(x,y) (18)
*#(x,s(y)) +#(x,*(x,y)) (19)
*#(x,s(y)) *#(x,y) (20)
f#(s(x),s(y)) f#(-(min(s(x),s(y)),max(s(x),s(y))),*(s(x),s(y))) (21)
f#(s(x),s(y)) -#(min(s(x),s(y)),max(s(x),s(y))) (22)
f#(s(x),s(y)) min#(s(x),s(y)) (23)
f#(s(x),s(y)) max#(s(x),s(y)) (24)
f#(s(x),s(y)) *#(s(x),s(y)) (25)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 6 components.