Certification Problem

Input (TPDB TRS_Standard/Secret_05_TRS/tpa1)

The rewrite relation of the following TRS is considered.

min(x,0) 0 (1)
min(0,y) 0 (2)
min(s(x),s(y)) s(min(x,y)) (3)
max(x,0) x (4)
max(0,y) y (5)
max(s(x),s(y)) s(max(x,y)) (6)
-(x,0) x (7)
-(s(x),s(y)) -(x,y) (8)
gcd(s(x),s(y)) gcd(-(s(max(x,y)),s(min(x,y))),s(min(x,y))) (9)

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) (10)
max#(s(x),s(y)) max#(x,y) (11)
-#(s(x),s(y)) -#(x,y) (12)
gcd#(s(x),s(y)) gcd#(-(s(max(x,y)),s(min(x,y))),s(min(x,y))) (13)
gcd#(s(x),s(y)) -#(s(max(x,y)),s(min(x,y))) (14)
gcd#(s(x),s(y)) max#(x,y) (15)
gcd#(s(x),s(y)) min#(x,y) (16)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 4 components.