Certification Problem

Input (TPDB TRS_Standard/AotoYamada_05/005)

The rewrite relation of the following TRS is considered.

app(app(plus,0),y) → y (1)
app(app(plus,app(s,x)),y) → app(s,app(app(plus,x),y)) (2)
app(app(app(curry,f),x),y) → app(app(f,x),y) (3)
add → app(curry,plus) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(add) = 5 weight(add) = 4
prec(curry) = 2 weight(curry) = 1
prec(s) = 1 weight(s) = 1
prec(app) = 0 weight(app) = 2
prec(0) = 7 weight(0) = 1
prec(plus) = 3 weight(plus) = 1
all of the following rules can be deleted.
app(app(plus,0),y) → y (1)
app(app(plus,app(s,x)),y) → app(s,app(app(plus,x),y)) (2)
app(app(app(curry,f),x),y) → app(app(f,x),y) (3)
add → app(curry,plus) (4)

1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.