Certification Problem

Input (TPDB TRS_Standard/SK90/4.54)

The rewrite relation of the following TRS is considered.

g(f(x,y)) f(f(g(g(x)),g(g(y))),f(g(g(x)),g(g(y)))) (1)

Property / Task

Prove or disprove termination.

Answer / Result

No.

Proof (by AProVE @ termCOMP 2023)

1 Innermost Lhss Increase

We add the following left hand sides to the innermost strategy.
g(f(x0,x1))

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
g#(f(x,y)) g#(g(x)) (2)
g#(f(x,y)) g#(x) (3)
g#(f(x,y)) g#(g(y)) (4)
g#(f(x,y)) g#(y) (5)
It remains to prove infiniteness of the resulting DP problem.

1.1.1 Narrowing Processor

We consider narrowings of the pair below position 1 to get the following set of pairs
g#(f(f(x0,x1),y1)) g#(f(f(g(g(x0)),g(g(x1))),f(g(g(x0)),g(g(x1))))) (6)

1.1.1.1 Narrowing Processor

We consider narrowings of the pair below position 1 to get the following set of pairs
g#(f(y0,f(x0,x1))) g#(f(f(g(g(x0)),g(g(x1))),f(g(g(x0)),g(g(x1))))) (7)

1.1.1.1.1 Instantiation Processor

The pairs are instantiated to the following pairs.
g#(f(x,y)) g#(y) (5)
g#(f(f(x0,x1),y1)) g#(f(f(g(g(x0)),g(g(x1))),f(g(g(x0)),g(g(x1))))) (6)
g#(f(y0,f(x0,x1))) g#(f(f(g(g(x0)),g(g(x1))),f(g(g(x0)),g(g(x1))))) (7)
g#(f(f(y_0,y_1),x1)) g#(f(y_0,y_1)) (8)
g#(f(f(f(y_0,y_1),y_2),x1)) g#(f(f(y_0,y_1),y_2)) (9)
g#(f(f(y_0,f(y_1,y_2)),x1)) g#(f(y_0,f(y_1,y_2))) (10)

1.1.1.1.1.1 Instantiation Processor

The pairs are instantiated to the following pairs.
g#(f(f(x0,x1),y1)) g#(f(f(g(g(x0)),g(g(x1))),f(g(g(x0)),g(g(x1))))) (6)
g#(f(y0,f(x0,x1))) g#(f(f(g(g(x0)),g(g(x1))),f(g(g(x0)),g(g(x1))))) (7)
g#(f(f(y_0,y_1),x1)) g#(f(y_0,y_1)) (8)
g#(f(f(f(y_0,y_1),y_2),x1)) g#(f(f(y_0,y_1),y_2)) (9)
g#(f(f(y_0,f(y_1,y_2)),x1)) g#(f(y_0,f(y_1,y_2))) (10)
g#(f(x0,f(y_0,y_1))) g#(f(y_0,y_1)) (11)
g#(f(x0,f(f(y_0,y_1),y_2))) g#(f(f(y_0,y_1),y_2)) (12)
g#(f(x0,f(y_0,f(y_1,y_2)))) g#(f(y_0,f(y_1,y_2))) (13)
g#(f(x0,f(f(f(y_0,y_1),y_2),y_3))) g#(f(f(f(y_0,y_1),y_2),y_3)) (14)
g#(f(x0,f(f(y_0,f(y_1,y_2)),y_3))) g#(f(f(y_0,f(y_1,y_2)),y_3)) (15)

1.1.1.1.1.1.1 Full Strategy Switch Processor

We have a locally confluent overlay TRS, no overlaps between P and R, and the strategy is less than innermost. Hence, it suffices to prove non-termination for the full rewrite relation.

Local Confluence Proof

All critical pairs are joinable within 0 step(s). 0

1.1.1.1.1.1.1.1 Loop

The following loop proves infiniteness of the DP problem.

t0 = g#(f(f(x0,x1),y1))
P g#(f(f(g(g(x0)),g(g(x1))),f(g(g(x0)),g(g(x1)))))
= t1
where t1 = t0σ and σ = {x0/g(g(x0)), x1/g(g(x1)), y1/f(g(g(x0)),g(g(x1)))}