Certification Problem

Input (TPDB TRS_Standard/AProVE_04/rta3)

The rewrite relation of the following TRS is considered.

ack(0,y) s(y) (1)
ack(s(x),0) ack(x,s(0)) (2)
ack(s(x),s(y)) ack(x,ack(s(x),y)) (3)
f(s(x),y) f(x,s(x)) (4)
f(x,s(y)) f(y,x) (5)
f(x,y) ack(x,y) (6)
ack(s(x),y) f(x,x) (7)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
ack#(s(x),0) ack#(x,s(0)) (8)
ack#(s(x),s(y)) ack#(s(x),y) (9)
ack#(s(x),s(y)) ack#(x,ack(s(x),y)) (10)
f#(s(x),y) f#(x,s(x)) (11)
f#(x,s(y)) f#(y,x) (12)
f#(x,y) ack#(x,y) (13)
ack#(s(x),y) f#(x,x) (14)

1.1 Size-Change Termination

Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

ack#(s(x),0) ack#(x,s(0)) (8)
1 > 1
ack#(s(x),s(y)) ack#(s(x),y) (9)
2 > 2
1 1
ack#(s(x),s(y)) ack#(x,ack(s(x),y)) (10)
1 > 1
f#(s(x),y) f#(x,s(x)) (11)
1 2
1 > 1
f#(x,s(y)) f#(y,x) (12)
2 > 1
1 2
f#(x,y) ack#(x,y) (13)
2 2
1 1
ack#(s(x),y) f#(x,x) (14)
1 > 2
1 > 1

As there is no critical graph in the transitive closure, there are no infinite chains.