Certification Problem

Input (TPDB TRS_Standard/SK90/2.42)

The rewrite relation of the following TRS is considered.

flatten(nil) → nil (1)
flatten(unit(x)) → flatten(x) (2)
flatten(++(x,y)) → ++(flatten(x),flatten(y)) (3)
flatten(++(unit(x),y)) → ++(flatten(x),flatten(y)) (4)
flatten(flatten(x)) → flatten(x) (5)
rev(nil) → nil (6)
rev(unit(x)) → unit(x) (7)
rev(++(x,y)) → ++(rev(y),rev(x)) (8)
rev(rev(x)) → x (9)
++(x,nil) → x (10)
++(nil,y) → y (11)
++(++(x,y),z) → ++(x,++(y,z)) (12)

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
[rev(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[flatten(x1)] =
1 0 0
0 1 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[++(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
[nil] =
1 0 0
0 0 0
0 0 0
[unit(x1)] =
1 1 0
0 1 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
++(x,nil) → x (10)
++(nil,y) → y (11)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[rev(x1)] =
1 0 0
1 1 0
1 1 1
· x1 +
0 0 0
1 0 0
1 0 0
[flatten(x1)] =
1 0 0
1 1 0
1 1 1
· x1 +
0 0 0
0 0 0
0 0 0
[++(x1, x2)] =
1 0 0
0 1 0
0 1 0
· x1 +
1 0 0
0 1 0
0 1 0
· x2 +
1 0 0
0 0 0
1 0 0
[nil] =
0 0 0
0 0 0
0 0 0
[unit(x1)] =
1 1 0
0 1 0
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
flatten(unit(x)) → flatten(x) (2)
flatten(++(unit(x),y)) → ++(flatten(x),flatten(y)) (4)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[rev(x1)] =
1 0 0
1 1 1
1 1 0
· x1 +
0 0 0
1 0 0
1 0 0
[flatten(x1)] =
1 1 1
1 0 0
1 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[++(x1, x2)] =
1 0 0
0 1 0
0 0 1
· x1 +
1 0 0
0 1 0
0 0 1
· x2 +
1 0 0
0 0 0
0 0 0
[nil] =
0 0 0
1 0 0
1 0 0
[unit(x1)] =
1 0 0
1 0 0
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
flatten(nil) → nil (1)
flatten(flatten(x)) → flatten(x) (5)

1.1.1.1 Rule Removal

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

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[rev(x1)] =
1 1 0
1 0 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[flatten(x1)] =
1 0 0
0 0 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[++(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
[unit(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
rev(unit(x)) → unit(x) (7)

1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[rev(x1)] =
1 1 1
0 1 0
1 0 1
· x1 +
1 0 0
0 0 0
0 0 0
[flatten(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[++(x1, x2)] =
1 0 0
0 0 0
0 1 1
· x1 +
1 0 0
0 0 0
0 1 1
· x2 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
rev(rev(x)) → x (9)

1.1.1.1.1.1.1 Rule Removal

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

1.1.1.1.1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(rev) = 1 weight(rev) = 0
prec(++) = 0 weight(++) = 0
all of the following rules can be deleted.
rev(++(x,y)) → ++(rev(y),rev(x)) (8)
++(++(x,y),z) → ++(x,++(y,z)) (12)

1.1.1.1.1.1.1.1.1 R is empty

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