YES
Confluence Proof
Confluence Proof
by Hakusan
Input
The rewrite relation of the following TRS is considered.
|
f(x) |
→ |
g(k(x)) |
|
f(x) |
→ |
a |
|
g(x) |
→ |
a |
|
k(a) |
→ |
k(k(a)) |
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 = a {ε}←f(y1)→ε g(k(y1)) = t can be joined as follows.
s
↔
t
-
The critical peak s = g(k(y1)) {ε}←f(y1)→ε a = t can be joined as follows.
s
↔
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(y1) |
→ |
a |
|
f(y1) |
→ |
g(k(y1)) |
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
| [a] |
= |
|
| [g(x1)] |
= |
· x1 +
|
| [f(x1)] |
= |
· x1 +
|
| [k(x1)] |
= |
· x1 +
|
the
rule
remains in R.
Moreover,
the
rules
|
f(x) |
→ |
g(k(x)) |
|
k(a) |
→ |
k(k(a)) |
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
| [a] |
= |
|
| [g(x1)] |
= |
· x1 +
|
| [f(x1)] |
= |
· x1 +
|
| [k(x1)] |
= |
· x1 +
|
all rules of R could be removed.
Moreover,
the
rule
remains 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