Certification Problem

Input (TPDB TRS_Standard/SK90/2.39)

The rewrite relation of the following TRS is considered.

rev(nil) → nil (1)
rev(.(x,y)) → ++(rev(y),.(x,nil)) (2)
car(.(x,y)) → x (3)
cdr(.(x,y)) → y (4)
null(nil) → true (5)
null(.(x,y)) → false (6)
++(nil,y) → y (7)
++(.(x,y),z) → .(x,++(y,z)) (8)

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
[car(x1)] =
1 0 1
1 1 0
0 1 0
· x1 +
1 0 0
1 0 0
0 0 0
[false] =
0 0 0
0 0 0
0 0 0
[rev(x1)] =
1 1 0
1 1 0
1 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 1
0 0 1
· x2 +
0 0 0
0 0 0
0 0 0
[cdr(x1)] =
1 1 0
1 1 1
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[nil] =
1 0 0
0 0 0
1 0 0
[true] =
0 0 0
0 0 0
0 0 0
[null(x1)] =
1 1 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[.(x1, x2)] =
1 1 0
0 0 1
1 0 0
· x1 +
1 0 0
0 1 0
0 0 1
· x2 +
1 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
car(.(x,y)) → x (3)
cdr(.(x,y)) → y (4)
null(nil) → true (5)
null(.(x,y)) → false (6)
++(nil,y) → y (7)

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
1 0 0
0 0 0
· x1 +
1 0 0
1 0 0
0 0 0
[++(x1, x2)] =
1 0 0
0 1 0
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[nil] =
0 0 0
1 0 0
0 0 0
[.(x1, x2)] =
1 0 1
1 1 0
1 0 0
· x1 +
1 0 1
0 0 0
0 1 0
· x2 +
1 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
rev(nil) → nil (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
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
1 0 0
[++(x1, x2)] =
1 0 0
0 1 0
0 0 0
· x1 +
1 0 0
0 0 0
0 1 1
· x2 +
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 0
· x1 +
1 1 0
0 0 0
0 0 0
· x2 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
rev(.(x,y)) → ++(rev(y),.(x,nil)) (2)

1.1.1.1 Rule Removal

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

1.1.1.1.1 R is empty

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