Certification Problem

Input (TPDB TRS_Standard/SK90/4.30)

The rewrite relation of the following TRS is considered.

f(nil) → nil (1)
f(.(nil,y)) → .(nil,f(y)) (2)
f(.(.(x,y),z)) → f(.(x,.(y,z))) (3)
g(nil) → nil (4)
g(.(x,nil)) → .(g(x),nil) (5)
g(.(x,.(y,z))) → g(.(.(x,y),z)) (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
[f(x1)] =
1 1 0
0 1 0
1 1 1
· x1 +
1 0 0
0 0 0
1 0 0
[g(x1)] =
1 0 1
1 1 1
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[nil] =
0 0 0
1 0 0
0 0 0
[.(x1, x2)] =
1 0 0
1 0 0
0 0 1
· x1 +
1 0 0
0 1 0
1 0 0
· x2 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
f(nil) → nil (1)

1.1 Rule Removal

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

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[f(x1)] =
1 0 0
0 0 0
1 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[g(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[nil] =
0 0 0
0 0 0
0 0 0
[.(x1, x2)] =
1 0 0
0 0 0
0 0 1
· x1 +
1 0 0
0 0 0
1 0 0
· x2 +
1 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
g(.(x,.(y,z))) → g(.(.(x,y),z)) (6)

1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(g) = 1 weight(g) = 2
prec(.) = 0 weight(.) = 0
prec(f) = 3 weight(f) = 2
prec(nil) = 2 weight(nil) = 1
all of the following rules can be deleted.
f(.(nil,y)) → .(nil,f(y)) (2)
f(.(.(x,y),z)) → f(.(x,.(y,z))) (3)
g(.(x,nil)) → .(g(x),nil) (5)

1.1.1.1.1 R is empty

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