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Show that $p \leftrightarrow q$ and $\sim p \leftrightarrow \sim q$ are logically equivalent.

We know that $p\leftrightarrow q\equiv (p\rightarrow q)\wedge(q\rightarrow p)$

$p\leftrightarrow q\equiv (\sim p\vee q)\wedge(\sim q\vee p)$

$p\leftrightarrow q\equiv (\sim p\wedge \sim q)\vee(\sim p\wedge p)\vee(q\wedge \sim q)\vee(q\wedge p)$

$p\leftrightarrow q\equiv (\sim p\wedge \sim q)\vee(F)\vee(F)\vee(q\wedge p)$

$p\leftrightarrow q\equiv (\sim p\wedge \sim q)\vee(p\wedge q)$

${\color{Red}{p\leftrightarrow q\equiv (p\wedge q)\vee(\sim p\wedge \sim q)} }$  ---------------$>(1)$

and     $\sim p\leftrightarrow \sim q\equiv (\sim p\rightarrow \sim q)\wedge(\sim q\rightarrow \sim p)$

$\sim p\leftrightarrow \sim q\equiv (\sim(\sim p)\vee \sim q)\wedge(\sim(\sim q)\vee \sim p)$

$\sim p\leftrightarrow \sim q\equiv (p\vee \sim q)\wedge (q\vee \sim p)$

$\sim p\leftrightarrow \sim q\equiv (p\wedge q)\vee (p\wedge \sim p )\vee(\sim q\wedge q)\vee (\sim q\wedge \sim p)$

$\sim p\leftrightarrow \sim q\equiv (p\wedge q)\vee (F )\vee(F)\vee (\sim q\wedge \sim p)$

$\sim p\leftrightarrow \sim q\equiv (p\wedge q)\vee (\sim q\wedge \sim p)$

${\color{Red}{\sim p\leftrightarrow \sim q\equiv (p\wedge q)\vee (\sim p\wedge \sim q)} }$  -----------------$>(2)$

Hence$,$from equation $(1)$ and $(2)$  ${\color{Red}{p\leftrightarrow q\equiv \sim p\leftrightarrow\sim q} }$   both are equivalent.

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Now we have some other Bi-implication equivalents.

$\sim( p\leftrightarrow q)\equiv\sim\left[(p\rightarrow q)\wedge(q\rightarrow p)\right]$

$\sim( p\leftrightarrow q)\equiv\sim\left[(\sim p\vee q)\wedge(\sim q\vee p)\right]$

$\sim( p\leftrightarrow q)\equiv\sim\left[(\sim p\wedge\sim q)\vee (\sim p\wedge p)\vee(q\wedge\sim q)\vee (q\wedge p)\right]$

$\sim( p\leftrightarrow q)\equiv\sim\left[(\sim p\wedge\sim q)\vee (F)\vee(F)\vee (q\wedge p)\right]$

$\sim( p\leftrightarrow q)\equiv\sim\left[(\sim p\wedge\sim q)\vee(q\wedge p)\right]$

$\sim( p\leftrightarrow q)\equiv \sim(\sim p\wedge\sim q)\wedge\sim(q\wedge p)$

$\sim( p\leftrightarrow q)\equiv \sim(\sim p)\vee\sim (\sim q)\wedge\sim q\vee\sim p$

$\sim( p\leftrightarrow q)\equiv (p\vee q)\wedge(\sim q\vee\sim p)$

$\sim( p\leftrightarrow q)\equiv (p\wedge\sim q)\vee(p\wedge\sim p)\vee(q\wedge\sim q)\vee(q\wedge\sim p)$​​​​​​​

$\sim( p\leftrightarrow q)\equiv (p\wedge\sim q)\vee(F)\vee(F)\vee(q\wedge\sim p)$​​​​​​​

$\sim( p\leftrightarrow q)\equiv (p\wedge\sim q)\vee(q\wedge\sim p)$​​​​​​​

${\color{Magenta}{\sim( p\leftrightarrow q)\equiv (p\wedge\sim q)\vee(\sim p\wedge q)\equiv p\oplus q} }$​​​​​​​  ---------------$>(3)$

and

$\sim( \sim p\leftrightarrow \sim q)\equiv\sim\left[(\sim p\rightarrow \sim q)\wedge(\sim q\rightarrow \sim p)\right]$

$\sim( \sim p\leftrightarrow\sim q)\equiv\sim\left[(\sim( \sim p)\vee\sim q)\wedge(\sim(\sim q)\vee\sim p)\right]$

$\sim(\sim p\leftrightarrow \sim q)\equiv\sim\left[( p\vee\sim q)\wedge(q\vee\sim p)\right]$

$\sim(\sim p\leftrightarrow \sim q)\equiv\sim( p\vee\sim q)\vee\sim(q\vee\sim p)$

$\sim(\sim p\leftrightarrow \sim q)\equiv(\sim p\wedge q)\vee(\sim q\wedge p)$

$\sim(\sim p\leftrightarrow \sim q)\equiv(\sim p\wedge q)\vee(p\wedge \sim q)$

${\color{Magenta}{\sim(\sim p\leftrightarrow \sim q)\equiv (p\wedge\sim q)\vee(\sim p\wedge q)\equiv p\oplus q} }$​​​​​​​  ---------------$>(4)$

and

$\sim p\leftrightarrow q\equiv (\sim p\rightarrow q)\wedge(q\rightarrow \sim p)$

$\sim p\leftrightarrow q\equiv (\sim(\sim p)\vee q)\wedge(\sim q\vee \sim p)$

$\sim p\leftrightarrow q\equiv (p\vee q)\wedge(\sim q\vee \sim p)$

$\sim p\leftrightarrow q\equiv (p\wedge\sim q)\vee (p\wedge\sim p)\vee(q\wedge \sim q)\vee(q\wedge\sim p)$

$\sim p\leftrightarrow q\equiv (p\wedge\sim q)\vee (F)\vee(F)\vee(q\wedge\sim p)$

$\sim p\leftrightarrow q\equiv (p\wedge\sim q)\vee(q\wedge\sim p)$

${\color{Magenta}{\sim p\leftrightarrow q\equiv (p\wedge\sim q)\vee(\sim p\wedge q)\equiv p\oplus q} }$​​​​​​​  ---------------$>(5)$

and

$p\leftrightarrow \sim q\equiv (p\rightarrow \sim q)\wedge(\sim q\rightarrow p)$

$p\leftrightarrow \sim q\equiv (\sim p\vee \sim q)\wedge(\sim(\sim q)\vee p)$

$p\leftrightarrow \sim q\equiv (\sim p\vee \sim q)\wedge(q\vee p)$

$p\leftrightarrow \sim q\equiv (\sim p\wedge q)\vee (\sim p\wedge p)\vee(\sim q\wedge q)\vee(\sim q\wedge p)$

$p\leftrightarrow \sim q\equiv (\sim p\wedge q)\vee (F)\vee(F)\vee(\sim q\wedge p)$

$p\leftrightarrow \sim q\equiv (\sim p\wedge q)\vee(\sim q\wedge p)$

$p\leftrightarrow\sim q\equiv (\sim p\wedge q)\vee(p\wedge \sim q)$

${\color{Magenta} {p\leftrightarrow\sim q\equiv (p\wedge \sim q)\vee (\sim p\wedge q)\equiv p\oplus q}}$​​​​​​​  ---------------$>(6)$

From the equation $(3),(4)$ and $(5)$ all are equivalents.

${\color{Magenta} {\sim( p\leftrightarrow q)\equiv\sim( \sim p\leftrightarrow \sim q)\equiv\sim p\leftrightarrow q\equiv p\leftrightarrow\sim q\equiv p\oplus q}}$

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$(OR)$

Truth table

 $p$ $q$ $\sim p$ $\sim q$ $\sim p\leftrightarrow\sim q$ $p\leftrightarrow q$ T T F F T T T F F T F F F T T F F F F F T T T T
edited
+1
Can you try without Truth Table?
0

@ sir

+1 vote
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+1 vote
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