YES
Confluence Proof
Confluence Proof
by Hakusan
Input
The rewrite relation of the following TRS is considered.
|
f(g(x)) |
→ |
h(g(x),g(x)) |
|
f(s(x)) |
→ |
h(s(x),s(x)) |
|
g(x) |
→ |
s(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(s(y1)) {1}←f(g(y1))→ε h(g(y1),g(y1)) = t can be joined as follows.
s
↔ h(s(y1),s(y1)) ↔ h(s(y1),g(y1)) ↔
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(g(y1)) |
→ |
f(s(y1)) |
|
f(g(y1)) |
→ |
h(g(y1),g(y1)) |
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(h) |
= |
0 |
|
stat(h) |
= |
lex
|
| prec(s) |
= |
0 |
|
stat(s) |
= |
lex
|
| prec(f) |
= |
2 |
|
stat(f) |
= |
lex
|
| prec(g) |
= |
3 |
|
stat(g) |
= |
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