Certification Problem

Input (TPDB TRS_Standard/AProVE_08/id_inc)

The rewrite relation of the following TRS is considered.

f(s(x)) → f(id_inc(c(x,x))) (1)
f(c(s(x),y)) → g(c(x,y)) (2)
g(c(s(x),y)) → g(c(y,x)) (3)
g(c(x,s(y))) → g(c(y,x)) (4)
g(c(x,x)) → f(x) (5)
id_inc(c(x,y)) → c(id_inc(x),id_inc(y)) (6)
id_inc(s(x)) → s(id_inc(x)) (7)
id_inc(0) → 0 (8)
id_inc(0) → s(0) (9)

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

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[g(x1)] =
1 0 0
0 0 0
1 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[f(x1)] =
1 1 0
1 1 0
1 1 1
· x1 +
1 0 0
0 0 0
1 0 0
[id_inc(x1)] =
1 0 0
0 1 1
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
[0] =
0 0 0
0 0 0
1 0 0
[s(x1)] =
1 0 0
1 1 1
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
[c(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
f(c(s(x),y)) → g(c(x,y)) (2)

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 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[id_inc(x1)] =
1 0 0
1 0 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[0] =
0 0 0
0 0 0
1 0 0
[s(x1)] =
1 0 0
0 0 0
1 0 0
· x1 +
0 0 0
0 0 0
1 0 0
[c(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
f(s(x)) → f(id_inc(c(x,x))) (1)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[id_inc(x1)] = 2 · x1 + 6
[0] = 2
[s(x1)] = 1 · x1 + 1
[c(x1, x2)] = 4 · x1 + 1 · x2 + 28
all of the following rules can be deleted.
id_inc(c(x,y)) → c(id_inc(x),id_inc(y)) (6)
id_inc(s(x)) → s(id_inc(x)) (7)
id_inc(0) → 0 (8)
id_inc(0) → s(0) (9)

1.1.1.1.1 R is empty

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