Certification Problem
Input (COPS 62)
We consider the TRS containing the following rules:
|
if(true,x,y) |
→ |
x |
(1) |
|
if(false,x,y) |
→ |
y |
(2) |
|
-(s(x),s(y)) |
→ |
-(x,y) |
(3) |
|
-(x,0) |
→ |
x |
(4) |
|
-(0,x) |
→ |
0 |
(5) |
|
<(s(x),s(y)) |
→ |
<(x,y) |
(6) |
|
<(0,s(x)) |
→ |
true |
(7) |
|
<(x,0) |
→ |
false |
(8) |
|
mod(0,y) |
→ |
0 |
(9) |
|
mod(x,s(y)) |
→ |
if(<(x,s(y)),x,mod(-(x,s(y)),s(y))) |
(10) |
|
mod(x,0) |
→ |
x |
(11) |
|
gcd(x,y) |
→ |
gcd(y,mod(x,y)) |
(12) |
|
gcd(x,0) |
→ |
x |
(13) |
|
gcd(0,x) |
→ |
x |
(14) |
The underlying signature is as follows:
{if/3, true/0, false/0, -/2, s/1, 0/0, </2, mod/2, gcd/2}Property / Task
Prove or disprove confluence.Answer / Result
Yes.Proof (by csi @ CoCo 2022)
1 Critical Pair Closing System
Confluence is proven using the following terminating critical-pair-closing-system R:
|
<(0,s(x)) |
→ |
true |
(7) |
|
if(true,x,y) |
→ |
x |
(1) |
|
gcd(0,x) |
→ |
x |
(14) |
|
mod(x,0) |
→ |
x |
(11) |
|
mod(0,y) |
→ |
0 |
(9) |
|
gcd(x,0) |
→ |
x |
(13) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [gcd(x1, x2)] |
= |
1 · x1 + 2 · x2 + 3 |
| [if(x1, x2, x3)] |
= |
1 · x1 + 2 · x2 + 2 · x3 + 0 |
| [s(x1)] |
= |
1 · x1 + 0 |
| [0] |
= |
0 |
| [true] |
= |
0 |
| [mod(x1, x2)] |
= |
2 · x1 + 2 · x2 + 4 |
| [<(x1, x2)] |
= |
1 · x1 + 1 · x2 + 1 |
all of the following rules can be deleted.
|
<(0,s(x)) |
→ |
true |
(7) |
|
gcd(0,x) |
→ |
x |
(14) |
|
mod(x,0) |
→ |
x |
(11) |
|
mod(0,y) |
→ |
0 |
(9) |
|
gcd(x,0) |
→ |
x |
(13) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [if(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 4 · x3 + 1 |
| [true] |
= |
0 |
all of the following rules can be deleted.
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.