Certification Problem
Input (TPDB TRS_Standard/SK90/2.32)
The rewrite relation of the following TRS is considered.
|
not(x) |
→ |
if(x,false,true) |
(1) |
|
and(x,y) |
→ |
if(x,y,false) |
(2) |
|
or(x,y) |
→ |
if(x,true,y) |
(3) |
|
implies(x,y) |
→ |
if(x,y,true) |
(4) |
|
=(x,x) |
→ |
true |
(5) |
|
=(x,y) |
→ |
if(x,y,not(y)) |
(6) |
|
if(true,x,y) |
→ |
x |
(7) |
|
if(false,x,y) |
→ |
y |
(8) |
|
if(x,x,if(x,false,true)) |
→ |
true |
(9) |
|
=(x,y) |
→ |
if(x,y,if(y,false,true)) |
(10) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [and(x1, x2)] |
= |
2 · x1 + 18 · x2 + 0 |
| [false] |
= |
0 |
| [if(x1, x2, x3)] |
= |
2 · x1 + 18 · x2 + 4 · x3 + 0 |
| [or(x1, x2)] |
= |
2 · x1 + 4 · x2 + 0 |
| [not(x1)] |
= |
3 · x1 + 1 |
| [=(x1, x2)] |
= |
2 · x1 + 30 · x2 + 8 |
| [implies(x1, x2)] |
= |
16 · x1 + 18 · x2 + 0 |
| [true] |
= |
0 |
all of the following rules can be deleted.
|
not(x) |
→ |
if(x,false,true) |
(1) |
|
=(x,x) |
→ |
true |
(5) |
|
=(x,y) |
→ |
if(x,y,not(y)) |
(6) |
|
=(x,y) |
→ |
if(x,y,if(y,false,true)) |
(10) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(implies) |
= |
2 |
|
weight(implies) |
= |
3 |
|
|
|
| prec(or) |
= |
0 |
|
weight(or) |
= |
4 |
|
|
|
| prec(and) |
= |
4 |
|
weight(and) |
= |
4 |
|
|
|
| prec(if) |
= |
1 |
|
weight(if) |
= |
2 |
|
|
|
| prec(true) |
= |
3 |
|
weight(true) |
= |
1 |
|
|
|
| prec(false) |
= |
5 |
|
weight(false) |
= |
2 |
|
|
|
all of the following rules can be deleted.
|
and(x,y) |
→ |
if(x,y,false) |
(2) |
|
or(x,y) |
→ |
if(x,true,y) |
(3) |
|
implies(x,y) |
→ |
if(x,y,true) |
(4) |
|
if(true,x,y) |
→ |
x |
(7) |
|
if(false,x,y) |
→ |
y |
(8) |
|
if(x,x,if(x,false,true)) |
→ |
true |
(9) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.