YES
Confluence Proof
Confluence Proof
by Hakusan
Input
The rewrite relation of the following TRS is considered.
|
+(0,y) |
→ |
y |
|
+(s(x),y) |
→ |
s(+(x,y)) |
|
+(x,0) |
→ |
x |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
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 = 0 {ε}←+(0,0)→ε 0 = t can be joined as follows.
s
↔
t
-
The critical peak s = s(+(0,x0_2)) {ε}←+(0,s(x0_2))→ε s(x0_2) = t can be joined as follows.
s
↔
t
-
The critical peak s = s(y1) {ε}←+(s(y1),0)→ε s(+(y1,0)) = t can be joined as follows.
s
↔
t
-
The critical peak s = s(+(s(y1),x0_2)) {ε}←+(s(y1),s(x0_2))→ε s(+(y1,s(x0_2))) = t can be joined as follows.
s
↔ s(s(+(y1,x0_2))) ↔
t
-
The critical peak s = 0 {ε}←+(0,0)→ε 0 = t can be joined as follows.
s
↔
t
-
The critical peak s = s(+(x0_1,0)) {ε}←+(s(x0_1),0)→ε s(x0_1) = t can be joined as follows.
s
↔
t
-
The critical peak s = s(y2) {ε}←+(0,s(y2))→ε s(+(0,y2)) = t can be joined as follows.
s
↔
t
-
The critical peak s = s(+(x0_1,s(y2))) {ε}←+(s(x0_1),s(y2))→ε s(+(s(x0_1),y2)) = t can be joined as follows.
s
↔ s(s(+(x0_1,y2))) ↔
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.
|
+(0,s(x0_2)) |
→ |
s(+(0,x0_2)) |
|
+(0,s(x0_2)) |
→ |
s(x0_2) |
|
+(s(y1),0) |
→ |
s(y1) |
|
+(s(y1),0) |
→ |
s(+(y1,0)) |
|
+(s(y1),s(x0_2)) |
→ |
s(+(s(y1),x0_2)) |
|
+(s(y1),s(x0_2)) |
→ |
s(+(y1,s(x0_2))) |
|
+(s(x0_1),0) |
→ |
s(+(x0_1,0)) |
|
+(s(x0_1),0) |
→ |
s(x0_1) |
|
+(0,s(y2)) |
→ |
s(y2) |
|
+(0,s(y2)) |
→ |
s(+(0,y2)) |
|
+(s(x0_1),s(y2)) |
→ |
s(+(x0_1,s(y2))) |
|
+(s(x0_1),s(y2)) |
→ |
s(+(s(x0_1),y2)) |
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(+) |
= |
1 |
|
stat(+) |
= |
lex
|
| prec(s) |
= |
0 |
|
stat(s) |
= |
lex
|
| prec(0) |
= |
0 |
|
stat(0) |
= |
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