Question Number 1689 by 123456 last updated on 31/Aug/15
$$\mathrm{if}\:{p}\:\mathrm{and}\:{q}\:\mathrm{are}\:\mathrm{two}\:\mathrm{afirmation} \\ $$$$\sim\left({p}\rightarrow{q}\right)=? \\ $$$$\sim\left({p}\leftarrow{q}\right)=? \\ $$$$\sim\left({p}\Leftrightarrow{q}\right)=? \\ $$
Answered by 112358 last updated on 01/Sep/15
![We will use the fact that p⇒q≡∽p∨q. Therefore, ∽(p⇒q)≡∽(∽p∨q) ≡∽(∽p)∧∽q De Morgan′s law ≡p∧∽q . Involution law Similary,∽(p←q)≡∽(q⇒p)≡q∧∽p . By definition, p⇔q≡(p⇒q)∧(q⇒p). ∴∽(p⇔q)≡∽[(p⇒q)∧(q⇒p)] ≡∽(p⇒q)∨∽(q⇒p) ≡(p∧∽q)∨(q∧∽p)](Q1699.png)
$${We}\:{will}\:{use}\:{the}\:{fact}\:{that}\: \\ $$$${p}\Rightarrow{q}\equiv\backsim{p}\vee{q}.\:{Therefore}, \\ $$$$\backsim\left({p}\Rightarrow{q}\right)\equiv\backsim\left(\backsim{p}\vee{q}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\equiv\backsim\left(\backsim{p}\right)\wedge\backsim{q}\:\:\:\:\:\:{De}\:{Morgan}'{s}\:{law} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\equiv{p}\wedge\backsim{q}\:\:\:\:\:\:\:\:\:\:\:\:.\:\:\:\:\:\:{Involution}\:{law} \\ $$$${Similary},\backsim\left({p}\leftarrow{q}\right)\equiv\backsim\left({q}\Rightarrow{p}\right)\equiv{q}\wedge\backsim{p}\:. \\ $$$${By}\:{definition}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{p}\Leftrightarrow{q}\equiv\left({p}\Rightarrow{q}\right)\wedge\left({q}\Rightarrow{p}\right). \\ $$$$\therefore\backsim\left({p}\Leftrightarrow{q}\right)\equiv\backsim\left[\left({p}\Rightarrow{q}\right)\wedge\left({q}\Rightarrow{p}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\equiv\backsim\left({p}\Rightarrow{q}\right)\vee\backsim\left({q}\Rightarrow{p}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\equiv\left({p}\wedge\backsim{q}\right)\vee\left({q}\wedge\backsim{p}\right) \\ $$$$ \\ $$$$ \\ $$