Certification Problem
Input (COPS 578)
We consider the TRS containing the following rules:
|
s(p(x)) |
→ |
x |
(1) |
|
p(s(x)) |
→ |
x |
(2) |
|
+(x,0) |
→ |
x |
(3) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(4) |
|
+(x,p(y)) |
→ |
p(+(x,y)) |
(5) |
|
-(0,0) |
→ |
0 |
(6) |
|
-(x,s(y)) |
→ |
p(-(x,y)) |
(7) |
|
-(x,p(y)) |
→ |
s(-(x,y)) |
(8) |
|
-(p(x),y) |
→ |
p(-(x,y)) |
(9) |
|
-(s(x),y) |
→ |
s(-(x,y)) |
(10) |
The underlying signature is as follows:
{s/1, p/1, +/2, 0/0, -/2}Property / Task
Prove or disprove confluence.Answer / Result
Yes.Proof (by csi @ CoCo 2023)
1 Redundant Rules Transformation
To prove that the TRS is (non-)confluent, we show (non-)confluence of the following
modified system:
|
-(s(x),y) |
→ |
s(-(x,y)) |
(10) |
|
-(p(x),y) |
→ |
p(-(x,y)) |
(9) |
|
-(x,p(y)) |
→ |
s(-(x,y)) |
(8) |
|
-(x,s(y)) |
→ |
p(-(x,y)) |
(7) |
|
-(0,0) |
→ |
0 |
(6) |
|
+(x,p(y)) |
→ |
p(+(x,y)) |
(5) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(4) |
|
+(x,0) |
→ |
x |
(3) |
|
p(s(x)) |
→ |
x |
(2) |
|
s(p(x)) |
→ |
x |
(1) |
All redundant rules that were added or removed can be
simulated in 2 steps
.
1.1 Critical Pair Closing System
Confluence is proven using the following terminating critical-pair-closing-system R:
|
-(x,p(y)) |
→ |
s(-(x,y)) |
(8) |
|
-(s(x),y) |
→ |
s(-(x,y)) |
(10) |
|
-(x,s(y)) |
→ |
p(-(x,y)) |
(7) |
|
s(p(x)) |
→ |
x |
(1) |
|
p(s(x)) |
→ |
x |
(2) |
|
-(p(x),y) |
→ |
p(-(x,y)) |
(9) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(4) |
|
+(x,p(y)) |
→ |
p(+(x,y)) |
(5) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [-(x1, x2)] |
= |
2 · x1 + 1 · x2 + 0 |
| [s(x1)] |
= |
1 · x1 + 1 |
| [+(x1, x2)] |
= |
4 · x1 + 1 · x2 + 0 |
| [p(x1)] |
= |
1 · x1 + 1 |
all of the following rules can be deleted.
|
-(s(x),y) |
→ |
s(-(x,y)) |
(10) |
|
s(p(x)) |
→ |
x |
(1) |
|
p(s(x)) |
→ |
x |
(2) |
|
-(p(x),y) |
→ |
p(-(x,y)) |
(9) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [-(x1, x2)] |
= |
4 · x1 + 4 · x2 + 0 |
| [s(x1)] |
= |
1 · x1 + 5 |
| [+(x1, x2)] |
= |
4 · x1 + 1 · x2 + 4 |
| [p(x1)] |
= |
1 · x1 + 2 |
all of the following rules can be deleted.
|
-(x,p(y)) |
→ |
s(-(x,y)) |
(8) |
|
-(x,s(y)) |
→ |
p(-(x,y)) |
(7) |
1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [s(x1)] |
= |
1 · x1 + 3 |
| [+(x1, x2)] |
= |
1 · x1 + 2 · x2 + 2 |
| [p(x1)] |
= |
1 · x1 + 0 |
all of the following rules can be deleted.
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(4) |
1.1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [+(x1, x2)] |
= |
1 · x1 + 4 · x2 + 0 |
| [p(x1)] |
= |
1 · x1 + 5 |
all of the following rules can be deleted.
|
+(x,p(y)) |
→ |
p(+(x,y)) |
(5) |
1.1.1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.