Certification Problem
Input (COPS 581)
We consider the TRS containing the following rules:
|
+(0,y) |
→ |
y |
(1) |
|
+(s(0),y) |
→ |
s(+(0,y)) |
(2) |
|
+(x,y) |
→ |
+(y,x) |
(3) |
|
s(s(x)) |
→ |
x |
(4) |
The underlying signature is as follows:
{+/2, 0/0, s/1}Property / Task
Prove or disprove confluence.Answer / Result
Yes.Proof (by csi @ CoCo 2021)
1 Redundant Rules Transformation
To prove that the TRS is (non-)confluent, we show (non-)confluence of the following
modified system:
|
s(s(x)) |
→ |
x |
(4) |
|
+(x,y) |
→ |
+(y,x) |
(3) |
|
+(s(0),y) |
→ |
s(+(0,y)) |
(2) |
|
+(0,y) |
→ |
y |
(1) |
|
+(y,0) |
→ |
y |
(5) |
|
+(y,s(0)) |
→ |
s(+(0,y)) |
(6) |
All redundant rules that were added or removed can be
simulated in 2 steps
.
1.1 Decreasing Diagrams
1.1.2 Rule Labeling
Confluence is proven, because all critical peaks can be joined decreasingly
using the following rule labeling function (rules that are not shown have label 0).
-
↦ 0
-
↦ 6
-
|
+(s(0),y) |
→ |
s(+(0,y)) |
(2) |
↦ 5
-
↦ 1
-
↦ 0
-
|
+(y,s(0)) |
→ |
s(+(0,y)) |
(6) |
↦ 0
The critical pairs can be joined as follows. Here,
↔ is always chosen as an appropriate rewrite relation which
is automatically inferred by the certifier.
-
The critical peak s = s(x140)←→ε s(x140) = t can be joined as follows.
s
↔
t
-
The critical peak s = +(y,s(0))←→ε s(+(0,y)) = t can be joined as follows.
s
↔
t
-
The critical peak s = +(y,0)←→ε y = t can be joined as follows.
s
↔
t
-
The critical peak s = +(0,y)←→ε y = t can be joined as follows.
s
↔
t
-
The critical peak s = +(s(0),y)←→ε s(+(0,y)) = t can be joined as follows.
s
↔
t
-
The critical peak s = s(+(0,y))←→ε +(y,s(0)) = t can be joined as follows.
s
↔ s(+(0,y)) ↔
t
-
The critical peak s = s(+(0,0))←→ε s(0) = t can be joined as follows.
s
↔
t
-
The critical peak s = s(+(0,s(0)))←→ε s(+(0,s(0))) = t can be joined as follows.
s
↔
t
-
The critical peak s = y←→ε +(y,0) = t can be joined as follows.
s
↔ y ↔
t
-
The critical peak s = 0←→ε 0 = t can be joined as follows.
s
↔
t
-
The critical peak s = s(0)←→ε s(+(0,0)) = t can be joined as follows.
s
↔ s(0) ↔
t
-
The critical peak s = x←→ε +(0,x) = t can be joined as follows.
s
↔ x ↔
t
-
The critical peak s = s(0)←→ε s(+(0,0)) = t can be joined as follows.
s
↔ s(0) ↔
t
-
The critical peak s = 0←→ε 0 = t can be joined as follows.
s
↔
t
-
The critical peak s = s(+(0,x))←→ε +(s(0),x) = t can be joined as follows.
s
↔ s(+(0,x)) ↔
t
-
The critical peak s = s(+(0,s(0)))←→ε s(+(0,s(0))) = t can be joined as follows.
s
↔
t
-
The critical peak s = s(+(0,0))←→ε s(0) = t can be joined as follows.
s
↔
t
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