Certification Problem

Input (TPDB TRS_Standard/Secret_07_TRS/7)

The rewrite relation of the following TRS is considered.

f(a,f(a,f(b,f(x,y)))) → f(b,f(c,f(b,f(a,f(a,f(a,f(x,y))))))) (1)
f(a,f(c,f(x,y))) → f(b,f(x,y)) (2)

Property / Task

Prove or disprove termination.

Answer / Result

No.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
f#(a,f(a,f(b,f(x,y)))) → f#(b,f(c,f(b,f(a,f(a,f(a,f(x,y))))))) (3)
f#(a,f(a,f(b,f(x,y)))) → f#(c,f(b,f(a,f(a,f(a,f(x,y)))))) (4)
f#(a,f(a,f(b,f(x,y)))) → f#(b,f(a,f(a,f(a,f(x,y))))) (5)
f#(a,f(a,f(b,f(x,y)))) → f#(a,f(a,f(a,f(x,y)))) (6)
f#(a,f(a,f(b,f(x,y)))) → f#(a,f(a,f(x,y))) (7)
f#(a,f(a,f(b,f(x,y)))) → f#(a,f(x,y)) (8)
f#(a,f(c,f(x,y))) → f#(b,f(x,y)) (9)
It remains to prove infiniteness of the resulting DP problem.

1.1 Pair and Rule Removal

Some pairs and rules have been removed and it remains to prove infiniteness of the remaing problem. The following pairs have been deleted.
f#(a,f(a,f(b,f(x,y)))) → f#(b,f(c,f(b,f(a,f(a,f(a,f(x,y))))))) (3)
f#(a,f(a,f(b,f(x,y)))) → f#(c,f(b,f(a,f(a,f(a,f(x,y)))))) (4)
f#(a,f(a,f(b,f(x,y)))) → f#(b,f(a,f(a,f(a,f(x,y))))) (5)
f#(a,f(c,f(x,y))) → f#(b,f(x,y)) (9)
and the following rules have been deleted.

1.1.1 Narrowing Processor

We consider narrowings of the pair below position 2 to get the following set of pairs
f#(a,f(a,f(b,f(b,f(x0,x1))))) → f#(a,f(b,f(c,f(b,f(a,f(a,f(a,f(x0,x1)))))))) (10)
f#(a,f(a,f(b,f(a,f(b,f(x0,x1)))))) → f#(a,f(a,f(b,f(c,f(b,f(a,f(a,f(a,f(x0,x1))))))))) (11)
f#(a,f(a,f(b,f(c,f(x0,x1))))) → f#(a,f(a,f(b,f(x0,x1)))) (12)
f#(a,f(a,f(b,f(a,f(a,f(b,f(x0,x1))))))) → f#(a,f(a,f(a,f(b,f(c,f(b,f(a,f(a,f(a,f(x0,x1)))))))))) (13)
f#(a,f(a,f(b,f(a,f(c,f(x0,x1)))))) → f#(a,f(a,f(a,f(b,f(x0,x1))))) (14)

1.1.1.1 Pair and Rule Removal

Some pairs and rules have been removed and it remains to prove infiniteness of the remaing problem. The following pairs have been deleted.
f#(a,f(a,f(b,f(b,f(x0,x1))))) → f#(a,f(b,f(c,f(b,f(a,f(a,f(a,f(x0,x1)))))))) (10)
and the following rules have been deleted.

1.1.1.1.1 Instantiation Processor

The pairs are instantiated to the following pairs.
f#(a,f(a,f(b,f(x,y)))) → f#(a,f(a,f(x,y))) (7)
f#(a,f(a,f(b,f(a,f(b,f(x0,x1)))))) → f#(a,f(a,f(b,f(c,f(b,f(a,f(a,f(a,f(x0,x1))))))))) (11)
f#(a,f(a,f(b,f(c,f(x0,x1))))) → f#(a,f(a,f(b,f(x0,x1)))) (12)
f#(a,f(a,f(b,f(a,f(a,f(b,f(x0,x1))))))) → f#(a,f(a,f(a,f(b,f(c,f(b,f(a,f(a,f(a,f(x0,x1)))))))))) (13)
f#(a,f(a,f(b,f(a,f(c,f(x0,x1)))))) → f#(a,f(a,f(a,f(b,f(x0,x1))))) (14)
f#(a,f(a,f(b,f(a,x1)))) → f#(a,f(a,x1)) (15)

1.1.1.1.1.1 Loop

The following loop proves infiniteness of the DP problem.

t0 = f#(a,f(a,f(a,f(a,f(a,f(c,f(b,f(x,y))))))))
→R f#(a,f(a,f(a,f(a,f(b,f(b,f(x,y)))))))
→R f#(a,f(a,f(b,f(c,f(b,f(a,f(a,f(a,f(b,f(x,y))))))))))
→R f#(a,f(a,f(b,f(c,f(b,f(a,f(b,f(c,f(b,f(a,f(a,f(a,f(x,y)))))))))))))
→P f#(a,f(a,f(b,f(b,f(a,f(b,f(c,f(b,f(a,f(a,f(a,f(x,y))))))))))))
→P f#(a,f(a,f(b,f(a,f(b,f(c,f(b,f(a,f(a,f(a,f(x,y)))))))))))
→P f#(a,f(a,f(b,f(c,f(b,f(a,f(a,f(a,f(c,f(b,f(a,f(a,f(a,f(x,y))))))))))))))
→P f#(a,f(a,f(b,f(b,f(a,f(a,f(a,f(c,f(b,f(a,f(a,f(a,f(x,y)))))))))))))
→P f#(a,f(a,f(b,f(a,f(a,f(a,f(c,f(b,f(a,f(a,f(a,f(x,y))))))))))))
→P f#(a,f(a,f(a,f(a,f(a,f(c,f(b,f(a,f(a,f(a,f(x,y)))))))))))
= t9
where t9 = t0σ and σ = {x/a, y/f(a,f(a,f(x,y)))}