YES
Confluence Proof
Confluence Proof
by Hakusan
Input
The rewrite relation of the following TRS is considered.
|
f(g(g(x))) |
→ |
a |
|
f(g(h(x))) |
→ |
b |
|
f(h(g(x))) |
→ |
b |
|
f(h(h(x))) |
→ |
c |
|
g(x) |
→ |
h(x) |
| a |
→ |
b |
| b |
→ |
c |
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(h(g(y1))) {1}←f(g(g(y1)))→ε a = t can be joined as follows.
s
↔ b ↔
t
-
The critical peak s = f(g(h(y1))) {1.1}←f(g(g(y1)))→ε a = t can be joined as follows.
s
↔ b ↔
t
-
The critical peak s = f(h(h(y1))) {1}←f(g(h(y1)))→ε b = t can be joined as follows.
s
↔ c ↔
t
-
The critical peak s = f(h(h(y1))) {1.1}←f(h(g(y1)))→ε b = t can be joined as follows.
s
↔ 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(g(g(y1))) |
→ |
f(h(g(y1))) |
|
f(g(g(y1))) |
→ |
a |
|
f(g(g(y1))) |
→ |
f(g(h(y1))) |
|
f(g(h(y1))) |
→ |
f(h(h(y1))) |
|
f(g(h(y1))) |
→ |
b |
|
f(h(g(y1))) |
→ |
f(h(h(y1))) |
|
f(h(g(y1))) |
→ |
b |
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(c) |
= |
0 |
|
stat(c) |
= |
lex
|
| prec(b) |
= |
1 |
|
stat(b) |
= |
lex
|
| prec(a) |
= |
4 |
|
stat(a) |
= |
lex
|
| prec(h) |
= |
4 |
|
stat(h) |
= |
lex
|
| prec(f) |
= |
0 |
|
stat(f) |
= |
lex
|
| prec(g) |
= |
4 |
|
stat(g) |
= |
lex
|
the
rules
|
f(g(g(y1))) |
→ |
f(h(g(y1))) |
|
f(g(g(y1))) |
→ |
f(g(h(y1))) |
|
f(g(h(y1))) |
→ |
f(h(h(y1))) |
|
f(h(g(y1))) |
→ |
f(h(h(y1))) |
remain in R.
Moreover,
the
rule
remains in S.
1.1.1 Rule Removal
Using the
recursive path order with the following precedence and status
| prec(h) |
= |
0 |
|
stat(h) |
= |
lex
|
| prec(f) |
= |
0 |
|
stat(f) |
= |
lex
|
| prec(g) |
= |
1 |
|
stat(g) |
= |
lex
|
all rules of R could be removed.
Moreover,
all rules of S could be removed.
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