YES
Confluence Proof
Confluence Proof
by Hakusan
Input
The rewrite relation of the following TRS is considered.
|
F(H(x),y) |
→ |
G(H(x)) |
|
H(I(x)) |
→ |
I(x) |
|
F(I(x),y) |
→ |
G(I(x)) |
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(I(x1_1),y2) {1}←F(H(I(x1_1)),y2)→ε G(H(I(x1_1))) = t can be joined as follows.
s
↔ G(I(x1_1)) ↔
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(H(I(x1_1)),y2) |
→ |
F(I(x1_1),y2) |
|
F(H(I(x1_1)),y2) |
→ |
G(H(I(x1_1))) |
Relative termination of P / R is proven as follows.
1.1 Rule Removal
Using the
recursive path order with the following precedence and status
| prec(G) |
= |
0 |
|
stat(G) |
= |
lex
|
| prec(F) |
= |
1 |
|
stat(F) |
= |
lex
|
| prec(H) |
= |
0 |
|
stat(H) |
= |
lex
|
| prec(I) |
= |
0 |
|
stat(I) |
= |
lex
|
all rules of R could be removed.
Moreover,
all rules of S could be removed.
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