fix math mistake with implies instead of logic and

content/from-0-to-crypto-by-projects/episode-2-proof-demonstration.md

@@ -25,7 +25,7 @@ We could then guess that the researchers thanks then that:
 
 ![image](/gogo-s-blog-cpe/from-0-to-crypto-by-projects/episode-2-proof-demonstration/theory-behind-type7-hash.png)
 
-The question is to proove that there exists a function $ rev(hashed) $ so that $ \forall plain [rev(H(plain)) = plain] $ then $ \forall x [x = H(plain)] $
+The question is to proove that there exists a function $ rev(hashed) $ so that $ \forall plain [rev(H(plain)) = plain] $.
 
 
 ## IV/ 1- solving the theorem finding a way to proove the case.
@@ -40,8 +40,17 @@ There a serveral various operations including:
 - doing an boolean exclusive logical or to a known password. 
   -  as each number exclusively logically set to logical or (xored) with itself has the final value of 0 and as 0 set to logical or with another number will return this number, ![image](/gogo-s-blog-cpe/from-0-to-crypto-by-projects/episode-2-proof-demonstration/reversing-exclusive-or.png) it follows that logically set to logical or to the hardcoded password one time will change the values but logically set to logical or a second time to the same hardcoded value will change it to the original value. See [boolean algebra](https://en.wikipedia.org/wiki/Exclusive_or#Definition), and see this schems provided with the tool name `cryptool-2`.
 
-All of these are reversables.
+All of these are equivalent to another operation.
 
-Then I decide to choose a proof in the form: as $ A \implies B \implies C $, then $ A \implies C $.
+Then I decide to choose a proof in the form:
+
+as: $ \forall h \in N[(h = p + x) \implies (p = h - x)] $
+
+and as: $ \forall h \in N[(h = p \oplus x) \implies (p = h \oplus x)] $
+
+and as:
+$ \forall h \in N[(h = p \ggg x) \implies (p = h \lll x)] $
+
+then, for any h in N, $ p = (h - x_1 \oplus x_2 \lll x_3) $.
 
 Let's check it out that [in this paper]( /gogo-s-blog-cpe/from-0-to-crypto-by-projects/episode-2-proof-demonstration/latex-reverse-type7.pdf )!

static/from-0-to-crypto-by-projects/episode-2-proof-demonstration/latex-reverse-type7.tex

@@ -64,7 +64,7 @@
 
 ## II/ exclusive or
 
-According to the boolean algebra about the exclusive logical or operation, $ \forall x [y = (x \oplus x) \implies (y = 0)] $. $ \\ $
+According to the boolean algebra about the exclusive logical or operation, $ \forall x [y = (x \oplus x) \implies (y = 0)] $. \\
 Then as $ xlat \oplus xlat = 0 $, and as $ p \oplus 0 = p $, we know that the original password $p = (xlat \oplus h)$. $ \\ $
 
 ## III/ rotating 4 first to 4 last bits
@@ -82,16 +82,16 @@
 if $ (p_r \land 0x0f < 0x0a) \implies (p_r \land 0x0f + 0x30)$ or else $ (p_r \land 0x0f > 0xa0) \implies (p_r \land 0x0f + 0x37) \\ $
 
 So if the out has the 4 four bits value so that: 
-$ x \in { x | (0xf0 & x) \leq 0xa0 } \implies y = x + 0x30 $ $\\$
+$ x \in \{ x | (0xf0 & x) \leq 0xa0 \} \implies y = x + 0x30 $ $\\$
 
 So if the out has the 4 four bits value so that: 
-$ x \in { x | (0xf0 & x) > 0xa0 } \implies y = x + 0x37 $ $\\$
+$ x \in \{ x | (0xf0 & x) > 0xa0 \} \implies y = x + 0x37 $ $\\$
 
 So if the out has the 4 four first bits value so that:
-$ x \in { x | (0x0f & x) \leq 0x0a } \implies y = x + 0x30 $ $\\$
+$ x \in \{ x | (0x0f & x) \leq 0x0a \} \implies y = x + 0x30 $ $\\$
 
 So if the out has the 4 four first bits value so that:
-$ x \in { x | (0x0f & x) > 0x0a } \implies y = x + 0x37 $ $\\$
+$ x \in \{ x | (0x0f & x) > 0x0a \} \implies y = x + 0x37 $ $\\$
 
 first byte: \\
   as $ 0xa0 < 0xf0 + 0x30 < y \\ $ 
@@ -104,19 +104,19 @@
   -2: $ \forall y \in H(x)[(x \in \{ x | 0x0a < x \}) \implies (y \in \{y | y < 0x4a\})] \\$
 
 
-Then for both of any subnumber:
-that $ \forall y = H(x), x \in \{ x | x \leq 0xa \} \implies y = x + 0x30 \\$
+Then for both of any subnumber: \\
+that $ \forall y = H(x), x \in \{ x | x \leq 0xa \} \implies y = x + 0x30 $ \\
 and that $ \forall y = H(x), x \in \{ x | x > 0xa \} \implies y = x + 0x37 \\$
 
 It follows: \\
 that $ \forall y = H(x)[(y \in \{ y | 0 < y \leq 0x0a + 0x30 \}) \implies (x = y - 0x30)] $ then $ 0 < x < 0x0a $ \\
 and that $ \forall y = H(x)[(y \in \{ y | 0 < y \leq 0x0a + 0x37 \}) \implies (x = y - 0x30)] $ then $ 0x0a \leq x $ $\\$
 
-# V /communtativity:
+# V/ communtativity:
 
 Addition, substraction and $ \oplus $ are commutative. $ \\ $
 
-# VI / proof
+# VI/ proof
 
 Then we have already proven each piece of the theorem so that:
 $hp = (d, s, f, d, ;, k, f, o, A, ,, ., i, y, e, w, r, k, l, d, J, K, D, H, S, U, B, s, g, v, c, a, 6, 9, 8, 3, 4, n, c, x , v) \implies (\forall x \in hp[0 \geq x 0 \geq 256 \implies x \in hp]) $ \\