Certification Problem

Input (TPDB TRS_Standard/SK90/4.47)

The rewrite relation of the following TRS is considered.

f(g(i(a,b,b'),c),d) → if(e,f(.(b,c),d'),f(.(b',c),d')) (1)
f(g(h(a,b),c),d) → if(e,f(.(b,g(h(a,b),c)),d),f(c,d')) (2)

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
[c] =
1 0 0
1 0 0
0 0 0
[e] =
0 0 0
0 0 0
0 0 0
[b] =
0 0 0
0 0 0
1 0 0
[i(x1, x2, x3)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
1 0 0
0 1 0
0 1 0
· x3 +
0 0 0
0 0 0
1 0 0
[g(x1, x2)] =
1 0 0
0 0 1
0 1 0
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[d'] =
0 0 0
0 0 0
1 0 0
[a] =
0 0 0
1 0 0
1 0 0
[f(x1, x2)] =
1 1 0
1 1 0
0 1 1
· x1 +
1 0 0
0 0 1
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[h(x1, x2)] =
1 0 0
0 0 0
0 1 1
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[if(x1, x2, x3)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
1 0 0
1 0 0
· x2 +
1 0 0
1 0 0
0 0 1
· x3 +
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 +
0 0 0
0 0 0
0 0 0
[d] =
0 0 0
0 0 0
0 0 0
[b'] =
0 0 0
1 0 0
1 0 0
all of the following rules can be deleted.
f(g(i(a,b,b'),c),d) → if(e,f(.(b,c),d'),f(.(b',c),d')) (1)

1.1 Rule Removal

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals
[c] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[e] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[b] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[g(x1, x2)] =
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x2 +
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[d'] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[a] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[f(x1, x2)] =
1 0 1 0 0
0 0 0 0 0
0 0 0 0 0
0 0 1 0 0
0 0 0 0 0
· x1 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x2 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[h(x1, x2)] =
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x2 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[if(x1, x2, x3)] =
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x2 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x3 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
[.(x1, x2)] =
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x2 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[d] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
all of the following rules can be deleted.
f(g(h(a,b),c),d) → if(e,f(.(b,g(h(a,b),c)),d),f(c,d')) (2)

1.1.1 R is empty

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