Certification Problem

Input (TPDB TRS_Standard/SK90/4.40)

The rewrite relation of the following TRS is considered.

f(f(f(a,x),y),z) → f(f(x,z),f(y,z)) (1)
f(f(b,x),y) → x (2)
f(c,y) → y (3)

Property / Task

Prove or disprove termination.

Answer / Result

No.

Proof (by AProVE @ termCOMP 2023)

1 Uncurrying

We uncurry the binary symbol f in combination with the following symbol map which also determines the applicative arities of these symbols.

a is mapped to a, a1(x1), a2(x1, x2), a3(x1, x2, x3)
b is mapped to b, b1(x1), b2(x1, x2)
c is mapped to c, c1(x1)


There are no uncurry rules.
No rules have to be added for the eta-expansion.

Uncurrying the rules and adding the uncurrying rules yields the new set of rules
a3(x,y,z) → f(f(x,z),f(y,z)) (10)
b2(x,y) → x (11)
c1(y) → y (12)
f(a,y1) → a1(y1) (4)
f(a1(x0),y1) → a2(x0,y1) (5)
f(a2(x0,x1),y1) → a3(x0,x1,y1) (6)
f(b,y1) → b1(y1) (7)
f(b1(x0),y1) → b2(x0,y1) (8)
f(c,y1) → c1(y1) (9)

1.1 Innermost Lhss Increase

We add the following left hand sides to the innermost strategy.
a3(x0,x1,x2)
b2(x0,x1)
c1(x0)
f(a,x0)
f(a1(x0),x1)
f(a2(x0,x1),x2)
f(b,x0)
f(b1(x0),x1)
f(c,x0)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a3#(x,y,z) → f#(f(x,z),f(y,z)) (13)
a3#(x,y,z) → f#(x,z) (14)
a3#(x,y,z) → f#(y,z) (15)
f#(a2(x0,x1),y1) → a3#(x0,x1,y1) (16)
f#(b1(x0),y1) → b2#(x0,y1) (17)
f#(c,y1) → c1#(y1) (18)
It remains to prove infiniteness of the resulting DP problem.

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#(b1(x0),y1) → b2#(x0,y1) (17)
f#(c,y1) → c1#(y1) (18)
and the following rules have been deleted.

1.1.1.1.1 Narrowing Processor

We consider narrowings of the pair below position 1 to get the following set of pairs
a3#(a,y1,x0) → f#(a1(x0),f(y1,x0)) (19)
a3#(a1(x0),y1,x1) → f#(a2(x0,x1),f(y1,x1)) (20)
a3#(a2(x0,x1),y1,x2) → f#(a3(x0,x1,x2),f(y1,x2)) (21)
a3#(b,y1,x0) → f#(b1(x0),f(y1,x0)) (22)
a3#(b1(x0),y1,x1) → f#(b2(x0,x1),f(y1,x1)) (23)
a3#(c,y1,x0) → f#(c1(x0),f(y1,x0)) (24)

1.1.1.1.1.1 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.
a3#(a,y1,x0) → f#(a1(x0),f(y1,x0)) (19)
a3#(b,y1,x0) → f#(b1(x0),f(y1,x0)) (22)
and the following rules have been deleted.

1.1.1.1.1.1.1 Rewriting Processor

We rewrite the right hand side of the pair resulting in
→

1.1.1.1.1.1.1.1 Rewriting Processor

We rewrite the right hand side of the pair resulting in
→

1.1.1.1.1.1.1.1.1 Rewriting Processor

We rewrite the right hand side of the pair resulting in
→

1.1.1.1.1.1.1.1.1.1 Instantiation Processor

The pairs are instantiated to the following pairs.
f#(a2(x0,x1),y1) → a3#(x0,x1,y1) (16)
a3#(x,y,z) → f#(y,z) (15)
a3#(a1(x0),y1,x1) → f#(a2(x0,x1),f(y1,x1)) (20)
a3#(a2(x0,x1),y1,x2) → f#(f(f(x0,x2),f(x1,x2)),f(y1,x2)) (25)
a3#(b1(x0),y1,x1) → f#(x0,f(y1,x1)) (26)
a3#(c,y1,x0) → f#(x0,f(y1,x0)) (27)
a3#(a2(y_0,y_1),x1,x2) → f#(a2(y_0,y_1),x2) (28)

1.1.1.1.1.1.1.1.1.1.1 Instantiation Processor

The pairs are instantiated to the following pairs.
f#(a2(x0,x1),y1) → a3#(x0,x1,y1) (16)
a3#(a1(x0),y1,x1) → f#(a2(x0,x1),f(y1,x1)) (20)
a3#(a2(x0,x1),y1,x2) → f#(f(f(x0,x2),f(x1,x2)),f(y1,x2)) (25)
a3#(b1(x0),y1,x1) → f#(x0,f(y1,x1)) (26)
a3#(c,y1,x0) → f#(x0,f(y1,x0)) (27)
a3#(a2(y_0,y_1),x1,x2) → f#(a2(y_0,y_1),x2) (28)
a3#(x0,a2(y_0,y_1),x2) → f#(a2(y_0,y_1),x2) (29)

1.1.1.1.1.1.1.1.1.1.1.1 Instantiation Processor

The pairs are instantiated to the following pairs.
a3#(a1(x0),y1,x1) → f#(a2(x0,x1),f(y1,x1)) (20)
a3#(a2(x0,x1),y1,x2) → f#(f(f(x0,x2),f(x1,x2)),f(y1,x2)) (25)
a3#(b1(x0),y1,x1) → f#(x0,f(y1,x1)) (26)
a3#(c,y1,x0) → f#(x0,f(y1,x0)) (27)
a3#(a2(y_0,y_1),x1,x2) → f#(a2(y_0,y_1),x2) (28)
a3#(x0,a2(y_0,y_1),x2) → f#(a2(y_0,y_1),x2) (29)
f#(a2(a1(y_0),x1),x2) → a3#(a1(y_0),x1,x2) (30)
f#(a2(a2(y_0,y_1),x1),x2) → a3#(a2(y_0,y_1),x1,x2) (31)
f#(a2(b1(y_0),x1),x2) → a3#(b1(y_0),x1,x2) (32)
f#(a2(c,x1),x2) → a3#(c,x1,x2) (33)
f#(a2(x0,a2(y_1,y_2)),x2) → a3#(x0,a2(y_1,y_2),x2) (34)

1.1.1.1.1.1.1.1.1.1.1.1.1 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

1.1.1.1.1.1.1.1.1.1.1.1.1.1 Loop

The following loop proves infiniteness of the DP problem.

t0 = a3#(c,c,f(c,a2(c,c)))
→R a3#(c,c,c1(a2(c,c)))
→R a3#(c,c,a2(c,c))
→P f#(a2(c,c),f(c,a2(c,c)))
→P a3#(c,c,f(c,a2(c,c)))
= t4
where t4 = t0