Certification Problem

Input (TPDB TRS_Relative/Mixed_relative_TRS/relsubst)

The relative rewrite relation R/S is considered where R is the following TRS

o(lambda(x),y) lambda(o(x,d(1,o(y,p)))) (1)
o(d(x,y),z) d(o(x,z),o(y,z)) (2)
o(o(x,y),z) o(x,o(y,z)) (3)
lambda(x) x (4)
o(x,y) x (5)
o(x,y) y (6)
d(x,y) x (7)
d(x,y) y (8)

and S is the following TRS.

o(x,y) d(x,y) (9)
o(x,y) d(y,x) (10)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the Weighted Path Order with the following precedence and status
prec(d) = 0 status(d) = [1, 2] list-extension(d) = Lex
prec(p) = 0 status(p) = [] list-extension(p) = Lex
prec(1) = 0 status(1) = [] list-extension(1) = Lex
prec(o) = 1 status(o) = [1, 2] list-extension(o) = Lex
prec(lambda) = 0 status(lambda) = [1] list-extension(lambda) = Lex
and the following Max-polynomial interpretation
[d(x1, x2)] = max(0, 0 + 1 · x1, 0 + 1 · x2)
[p] = 0
[1] = 0
[o(x1, x2)] = 0 + 1 · x1 + 1 · x2
[lambda(x1)] = max(4, 4 + 1 · x1)
all of the following rules can be deleted.
o(lambda(x),y) lambda(o(x,d(1,o(y,p)))) (1)
o(d(x,y),z) d(o(x,z),o(y,z)) (2)
o(o(x,y),z) o(x,o(y,z)) (3)
lambda(x) x (4)
o(x,y) x (5)
o(x,y) y (6)
d(x,y) x (7)
d(x,y) y (8)
o(x,y) d(x,y) (9)
o(x,y) d(y,x) (10)

1.1 R is empty

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