Certification Problem

Input (TPDB TRS_Standard/AotoYamada_05/023)

The rewrite relation of the following TRS is considered.

app(id,x) → x (1)
app(plus,0) → id (2)
app(app(plus,app(s,x)),y) → app(s,app(app(plus,x),y)) (3)

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(s) = 0 weight(s) = 4
prec(0) = 3 weight(0) = 1
prec(plus) = 1 weight(plus) = 1
prec(app) = 5 weight(app) = 0
prec(id) = 4 weight(id) = 2
all of the following rules can be deleted.
app(id,x) → x (1)
app(plus,0) → id (2)
app(app(plus,app(s,x)),y) → app(s,app(app(plus,x),y)) (3)

1.1 R is empty

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