YES
Confluence Proof
Confluence Proof
by Hakusan
Input
The rewrite relation of the following TRS is considered.
| a |
→ |
c |
| b |
→ |
c |
|
f(a,b) |
→ |
d |
|
f(x,c) |
→ |
f(c,c) |
|
f(c,x) |
→ |
f(c,c) |
| d |
→ |
f(a,c) |
| d |
→ |
f(c,b) |
Proof
1 Compositional Parallel Critical Pair Systems
All parallel critical pairs of the TRS R are joinable by R.
This can be seen as follows:
The parallel 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 = f(c,c) {1, 2}←f(a,b)→ε d = t can be joined as follows.
s
↔ f(a,c) ↔
t
-
The critical peak s = f(c,b) {1}←f(a,b)→ε d = t can be joined as follows.
s
↔
t
-
The critical peak s = f(a,c) {2}←f(a,b)→ε d = t can be joined as follows.
s
↔
t
-
The critical peak s = f(c,c) {ε}←f(c,c)→ε f(c,c) = t can be joined as follows.
s
↔
t
-
The critical peak s = f(c,c) {ε}←f(c,c)→ε f(c,c) = t can be joined as follows.
s
↔
t
-
The critical peak s = f(c,b) {ε}←d→ε f(a,c) = t can be joined as follows.
s
↔ f(c,c) ↔
t
-
The critical peak s = f(a,c) {ε}←d→ε f(c,b) = t can be joined as follows.
s
↔ f(c,c) ↔
t
The TRS C is chosen as:
There are no rules.
Consequently, PCPS(R,C) is included in the following TRS P where
steps are used to show that certain pairs are C-convertible.
|
f(a,b) |
→ |
f(c,c) |
|
f(a,b) |
→ |
d |
|
f(a,b) |
→ |
f(c,b) |
|
f(a,b) |
→ |
f(a,c) |
| d |
→ |
f(c,b) |
| d |
→ |
f(a,c) |
Relative termination of P / R is proven as follows.
1.1 Rule Removal
Using the
linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1
over the naturals
| [d] |
= |
|
| [b] |
= |
|
| [c] |
= |
|
| [a] |
= |
|
| [f(x1, x2)] |
= |
· x1 + · x2 +
|
the
rules
remain in R.
Moreover,
the
rules
| a |
→ |
c |
| b |
→ |
c |
|
f(x,c) |
→ |
f(c,c) |
|
f(c,x) |
→ |
f(c,c) |
| d |
→ |
f(a,c) |
| d |
→ |
f(c,b) |
remain in S.
1.1.1 Rule Removal
Using the
linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1
over the naturals
| [d] |
= |
|
| [b] |
= |
|
| [c] |
= |
|
| [a] |
= |
|
| [f(x1, x2)] |
= |
· x1 + · x2 +
|
all rules of R could be removed.
Moreover,
the
rules
| a |
→ |
c |
| b |
→ |
c |
|
f(x,c) |
→ |
f(c,c) |
|
f(c,x) |
→ |
f(c,c) |
remain in S.
1.1.1.1 R is empty
There are no rules in the TRS R. Hence, R/S is relative terminating.
Confluence of C is proven as follows.
1.2 (Weakly) Orthogonal
Confluence is proven since the TRS is (weakly) orthogonal.
Tool configuration
Hakusan