Certification Problem                    
                
Input (COPS 138)
We consider the TRS containing the following rules:
| 
+(0,y) | 
→ | 
y | 
(1) | 
| 
+(x,0) | 
→ | 
x | 
(2) | 
| 
+(s(x),y) | 
→ | 
s(+(x,y)) | 
(3) | 
| 
+(x,s(y)) | 
→ | 
s(+(x,y)) | 
(4) | 
| 
+(x,+(y,z)) | 
→ | 
+(+(x,y),z) | 
(5) | 
| 
+(+(x,y),z) | 
→ | 
+(x,+(y,z)) | 
(6) | 
The underlying signature is as follows:
{+/2, 0/0, s/1}Property / Task
Prove or disprove confluence.Answer / Result
Yes.Proof (by csi @ CoCo 2022)
1 Redundant Rules Transformation
      To prove that the TRS is (non-)confluent, we show (non-)confluence of the following
      modified system:
      
| 
+(0,y) | 
→ | 
y | 
(1) | 
| 
+(x,0) | 
→ | 
x | 
(2) | 
| 
+(s(x),y) | 
→ | 
s(+(x,y)) | 
(3) | 
| 
+(x,s(y)) | 
→ | 
s(+(x,y)) | 
(4) | 
| 
+(+(x,y),z) | 
→ | 
+(x,+(y,z)) | 
(6) | 
      All redundant rules that were added or removed can be
      simulated in 4 steps
      .
1.1 Locally confluent and terminating
        Confluence is proven by showing local confluence and termination.
        1.1.1 Rule Removal
      Using the
      linear polynomial interpretation over the naturals
| [+(x1, x2)] | 
 =  | 
3 · x1 + 1 · x2 + 0 | 
| [0] | 
 =  | 
2 | 
| [s(x1)] | 
 =  | 
1 · x1 + 0 | 
                
      all of the following rules can be deleted.  
      
| 
+(0,y) | 
→ | 
y | 
(1) | 
| 
+(x,0) | 
→ | 
x | 
(2) | 
1.1.1.1 Rule Removal
      Using the
      linear polynomial interpretation over the naturals
| [+(x1, x2)] | 
 =  | 
4 · x1 + 1 · x2 + 0 | 
| [s(x1)] | 
 =  | 
1 · x1 + 2 | 
                
      all of the following rules can be deleted.  
      
| 
+(s(x),y) | 
→ | 
s(+(x,y)) | 
(3) | 
1.1.1.1.1 Rule Removal
      Using the
      linear polynomial interpretation over the naturals
| [+(x1, x2)] | 
 =  | 
2 · x1 + 1 · x2 + 5 | 
| [s(x1)] | 
 =  | 
1 · x1 + 4 | 
                
      all of the following rules can be deleted.  
      
| 
+(+(x,y),z) | 
→ | 
+(x,+(y,z)) | 
(6) | 
1.1.1.1.1.1 Rule Removal
      Using the
      linear polynomial interpretation over the naturals
| [+(x1, x2)] | 
 =  | 
1 · x1 + 4 · x2 + 0 | 
| [s(x1)] | 
 =  | 
1 · x1 + 5 | 
                
      all of the following rules can be deleted.  
      
| 
+(x,s(y)) | 
→ | 
s(+(x,y)) | 
(4) | 
1.1.1.1.1.1.1 R is empty 
There are no rules in the TRS. Hence, it is terminating.
1.1.2 Local Confluence Proof
                All critical pairs are joinable which can be seen by computing normal forms of all critical pairs.