Find the mistake in following proof With condition $x\neq c$
$\neg(\underbrace{0<x-c}_{p}) \Leftrightarrow \underbrace{0<c-x}_{\neg p}$
$ 1.(\underbrace{0<x-c}_{p} \vee \underbrace{0<c-x}_{\neg p})\wedge (\underbrace{\delta >x-c}_{q} \wedge \underbrace{\delta>c-x}_{r}))$
by distribution of conjunction over disjunction we have
$\Leftrightarrow(0<x-c\wedge\delta>x-c\wedge\delta>c-x)\vee(0<c-x\wedge\delta>x-c\wedge\delta>c-x)$
$\Leftrightarrow (c-\delta<x<c+\delta \wedge c<x)\vee(c-\delta<x<c+\delta \wedge x<c)$
$\Leftrightarrow \underbrace{c-\delta<x<c+\delta \wedge x\neq c}_{s}$
$ 2.(\underbrace{0<x-c}_{p}\wedge \underbrace{\delta >x-c}_{q})\vee(\underbrace{\delta>c-x}_{r}\wedge\underbrace{0<c-x}_{\neg p})$
$\Leftrightarrow (0<x-c<\delta)\vee (0<c-x<\delta)$
$\Leftrightarrow (-\delta<x-c<\delta \wedge 0<x-c)\vee(-\delta<c-x<\delta \wedge 0<c-x)$
$\Leftrightarrow (c-\delta<x<c+\delta \wedge c<x)\vee(c-\delta<x<c+\delta \wedge x<c)$
$\Leftrightarrow \underbrace{c-\delta<x<c+\delta \wedge x\neq c}_{s}$
Since $(p \vee \neg p)\wedge(q\wedge r)\not \Leftrightarrow (p\wedge q)\vee(r \wedge \neg p)$
Have $s \not \Leftrightarrow s$
 A: You are using propositional variables $p$, $q$, and $r$. As propositional variables, they have nothing logically to do with each other, and hence it is indeed true that $(p \lor \neg p) \land (q\land r)$ is not logically equivalent to $(p \land q) \lor (r \land \neg p)$
However, in your derivation you gave $p$, $q$, and $r$ a specific meaning and, moreover, a meaning that allowed you to combine some of the information contained therein. That is, you appealed to the internal structure of the statements, rather than purely $p$'s, $q$'s, and $r$'s. This meant that you seemed to monstrate a lgical equivalence that really wasn't.
Indeed, you were using all kinds of mathematical assumptions regarding the nature of numbers, and the use of addition, smaller than, etc. As such, all you really showed were certain mathematical equivalences, rather than logical ones.
Consider a simpler example:
$p$ is the claim $a<b$
$q$ is the claim $b>a$
Well, obviously we have $a<b \Leftrightarrow b>a$ ... and yet it is not a logical truth that $p \Leftrightarrow q$
