Certification Problem

Input (TPDB TRS_Standard/Zantema_05/z10)

The rewrite relation of the following TRS is considered.

a(lambda(x),y) lambda(a(x,p(1,a(y,t)))) (1)
a(p(x,y),z) p(a(x,z),a(y,z)) (2)
a(a(x,y),z) a(x,a(y,z)) (3)
a(id,x) x (4)
a(1,id) 1 (5)
a(t,id) t (6)
a(1,p(x,y)) x (7)
a(t,p(x,y)) y (8)

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(id) = 0 status(id) = [] list-extension(id) = Lex
prec(p) = 0 status(p) = [1, 2] list-extension(p) = Lex
prec(t) = 0 status(t) = [] list-extension(t) = Lex
prec(1) = 0 status(1) = [] list-extension(1) = Lex
prec(a) = 4 status(a) = [1, 2] list-extension(a) = Lex
prec(lambda) = 0 status(lambda) = [1] list-extension(lambda) = Lex
and the following Max-polynomial interpretation
[id] = max(4)
[p(x1, x2)] = max(0, 0 + 1 · x1, 0 + 1 · x2)
[t] = 0
[1] = 0
[a(x1, x2)] = 0 + 1 · x1 + 1 · x2
[lambda(x1)] = max(0, 2 + 1 · x1)
all of the following rules can be deleted.
a(lambda(x),y) lambda(a(x,p(1,a(y,t)))) (1)
a(p(x,y),z) p(a(x,z),a(y,z)) (2)
a(a(x,y),z) a(x,a(y,z)) (3)
a(id,x) x (4)
a(1,id) 1 (5)
a(t,id) t (6)
a(1,p(x,y)) x (7)
a(t,p(x,y)) y (8)

1.1 R is empty

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