Certification Problem
Input (TPDB TRS_Standard/SK90/2.44)
The rewrite relation of the following TRS is considered.
|
del(.(x,.(y,z))) |
→ |
f(=(x,y),x,y,z) |
(1) |
|
f(true,x,y,z) |
→ |
del(.(y,z)) |
(2) |
|
f(false,x,y,z) |
→ |
.(x,del(.(y,z))) |
(3) |
|
=(nil,nil) |
→ |
true |
(4) |
|
=(.(x,y),nil) |
→ |
false |
(5) |
|
=(nil,.(y,z)) |
→ |
false |
(6) |
|
=(.(x,y),.(u,v)) |
→ |
and(=(x,u),=(y,v)) |
(7) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1
over the naturals
| [true] |
= |
|
| [v] |
= |
|
| [del(x1)] |
= |
· x1 +
|
| [f(x1,...,x4)] |
= |
· x1 + · x2 + · x3 + · x4 +
|
| [false] |
= |
|
| [.(x1, x2)] |
= |
· x1 + · x2 +
|
| [u] |
= |
|
| [and(x1, x2)] |
= |
· x1 + · x2 +
|
| [nil] |
= |
|
| [=(x1, x2)] |
= |
· x1 + · x2 +
|
all of the following rules can be deleted.
|
del(.(x,.(y,z))) |
→ |
f(=(x,y),x,y,z) |
(1) |
|
f(true,x,y,z) |
→ |
del(.(y,z)) |
(2) |
|
f(false,x,y,z) |
→ |
.(x,del(.(y,z))) |
(3) |
|
=(nil,nil) |
→ |
true |
(4) |
|
=(.(x,y),nil) |
→ |
false |
(5) |
|
=(nil,.(y,z)) |
→ |
false |
(6) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(and) |
= |
0 |
|
weight(and) |
= |
6 |
|
|
|
| prec(v) |
= |
7 |
|
weight(v) |
= |
1 |
|
|
|
| prec(u) |
= |
3 |
|
weight(u) |
= |
1 |
|
|
|
| prec(=) |
= |
1 |
|
weight(=) |
= |
2 |
|
|
|
| prec(.) |
= |
2 |
|
weight(.) |
= |
4 |
|
|
|
all of the following rules can be deleted.
|
=(.(x,y),.(u,v)) |
→ |
and(=(x,u),=(y,v)) |
(7) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.