Certification Problem
Input (COPS 22)
We consider the TRS containing the following rules:
|
g(a) |
→ |
f(g(a)) |
(1) |
|
g(b) |
→ |
c(a) |
(2) |
| a |
→ |
b |
(3) |
|
f(x) |
→ |
h(x) |
(4) |
|
h(x) |
→ |
c(b) |
(5) |
The underlying signature is as follows:
{g/1, a/0, f/1, b/0, c/1, h/1}Property / Task
Prove or disprove confluence.Answer / Result
Yes.Proof (by csi @ CoCo 2020)
1 Redundant Rules Transformation
To prove that the TRS is (non-)confluent, we show (non-)confluence of the following
modified system:
|
h(x) |
→ |
c(b) |
(5) |
|
f(x) |
→ |
h(x) |
(4) |
| a |
→ |
b |
(3) |
|
g(b) |
→ |
c(a) |
(2) |
|
g(a) |
→ |
f(g(a)) |
(1) |
|
g(b) |
→ |
c(b) |
(6) |
|
f(g(a)) |
→ |
c(b) |
(7) |
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
-
↦ 0
-
↦ 3
-
↦ 0
-
↦ 5
-
↦ 4
-
↦ 1
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 = h(g(a))←→ε c(b) = t can be joined as follows.
s
↔
t
-
The critical peak s = g(b)←→ε f(g(a)) = t can be joined as follows.
s
↔ c(b) ↔
t
-
The critical peak s = f(g(b))←→ε c(b) = t can be joined as follows.
s
↔ h(g(b)) ↔
t
-
The critical peak s = c(a)←→ε c(b) = t can be joined as follows.
s
↔
t
-
The critical peak s = f(f(g(a)))←→ε c(b) = t can be joined as follows.
s
↔ h(f(g(a))) ↔
t
-
The critical peak s = c(b)←→ε c(a) = t can be joined as follows.
s
↔ c(b) ↔
t
-
The critical peak s = c(b)←→ε h(g(a)) = t can be joined as follows.
s
↔ c(b) ↔
t
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