Certification Problem

Input (TPDB TRS_Standard/Zantema_05/z29)

The rewrite relation of the following TRS is considered.

a(lambda(x),y) lambda(a(x,1)) (1)
a(lambda(x),y) lambda(a(x,a(y,t))) (2)
a(a(x,y),z) a(x,a(y,z)) (3)
lambda(x) x (4)
a(x,y) x (5)
a(x,y) y (6)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[a(x1, x2)] =
1 0 0
0 1 0
0 0 1
· x1 +
1 0 0
0 1 0
0 0 1
· x2 +
0 0 0
0 0 0
0 0 0
[t] =
0 0 0
0 0 0
0 0 0
[lambda(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
[1] =
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
lambda(x) x (4)

1.1 Rule Removal

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

1.1.1 R is empty

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