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 |
|
f(h(k1(a),k2(b))) |
→ |
f(h(c,d)) |
|
f(h(c,k2(b))) |
→ |
f(h(c,d)) |
|
f(h(k1(a),d)) |
→ |
f(h(c,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
↔
t
-
The critical peak s = f(h(c,d)) {1.1, 1.2}←f(h(k1(a),k2(b)))→ε f(h(c,d)) = t can be joined as follows.
s
↔
t
-
The critical peak s = f(h(c,k2(b))) {1.1}←f(h(k1(a),k2(b)))→ε f(h(c,d)) = t can be joined as follows.
s
↔
t
-
The critical peak s = f(h(k1(a),d)) {1.2}←f(h(k1(a),k2(b)))→ε f(h(c,d)) = t can be joined as follows.
s
↔
t
-
The critical peak s = f(h(c,d)) {1.2}←f(h(c,k2(b)))→ε f(h(c,d)) = t can be joined as follows.
s
↔
t
-
The critical peak s = f(h(c,d)) {1.1}←f(h(k1(a),d))→ε f(h(c,d)) = 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(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)) |
|
f(h(k1(a),k2(b))) |
→ |
f(h(c,k2(b))) |
|
f(h(k1(a),k2(b))) |
→ |
f(h(c,d)) |
|
f(h(k1(a),k2(b))) |
→ |
f(h(k1(a),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) |
= |
8 |
|
stat(k1) |
= |
lex
|
| prec(g) |
= |
9 |
|
stat(g) |
= |
lex
|
| prec(b) |
= |
0 |
|
stat(b) |
= |
lex
|
| prec(a) |
= |
1 |
|
stat(a) |
= |
lex
|
| prec(f) |
= |
0 |
|
stat(f) |
= |
lex
|
| prec(h) |
= |
2 |
|
stat(h) |
= |
lex
|
| prec(d) |
= |
0 |
|
stat(d) |
= |
lex
|
| prec(c) |
= |
1 |
|
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