YES
Confluence Proof
Confluence Proof
by Hakusan
Input
The rewrite relation of the following TRS is considered.
|
F(G(x,A,B)) |
→ |
x |
|
G(F(H(C,D)),x,y) |
→ |
H(K1(x),K2(y)) |
|
K1(A) |
→ |
C |
|
K2(B) |
→ |
D |
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(K1(A),K2(B))) {1}←F(G(F(H(C,D)),A,B))→ε F(H(C,D)) = t can be joined as follows.
s
↔ F(H(C,K2(B))) ↔
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(F(H(C,D)),A,B)) |
→ |
F(H(K1(A),K2(B))) |
|
F(G(F(H(C,D)),A,B)) |
→ |
F(H(C,D)) |
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(K2) |
= |
0 |
|
stat(K2) |
= |
lex
|
| prec(K1) |
= |
0 |
|
stat(K1) |
= |
lex
|
| prec(G) |
= |
9 |
|
stat(G) |
= |
lex
|
| prec(B) |
= |
8 |
|
stat(B) |
= |
lex
|
| prec(A) |
= |
0 |
|
stat(A) |
= |
lex
|
| prec(F) |
= |
0 |
|
stat(F) |
= |
lex
|
| prec(H) |
= |
0 |
|
stat(H) |
= |
lex
|
| prec(D) |
= |
0 |
|
stat(D) |
= |
lex
|
| prec(C) |
= |
0 |
|
stat(C) |
= |
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