Question Number 198750 by depressiveshrek last updated on 24/Oct/23
![Prove the following is a tautology: [(p⊻q)∧(p⇒r)]⇒(q⊻r)](https://www.tinkutara.com/question/Q198750.png)
$$\mathrm{Prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{a}\:\mathrm{tautology}: \\ $$$$\left[\left({p}\veebar{q}\right)\wedge\left({p}\Rightarrow{r}\right)\right]\Rightarrow\left({q}\veebar{r}\right) \\ $$
Answered by MathematicalUser2357 last updated on 29/Dec/23
![Only know until =∼[(p⊻q)∧(∼p∨r)]∨(q⊻r) =(p⊻^− q)∨p∧∼r∨(q⊻r) =∼p∧∼q∨p∧q∨p∧∼r∨∼p∧r =p∨q∨r∧(∼p∨∼r)](https://www.tinkutara.com/question/Q202560.png)
$$\mathrm{Only}\:\mathrm{know}\:\mathrm{until} \\ $$$$=\sim\left[\left({p}\veebar{q}\right)\wedge\left(\sim{p}\vee{r}\right)\right]\vee\left({q}\veebar{r}\right) \\ $$$$=\left({p}\overset{−} {\veebar}{q}\right)\vee{p}\wedge\sim{r}\vee\left({q}\veebar{r}\right) \\ $$$$=\sim{p}\wedge\sim{q}\vee{p}\wedge{q}\vee{p}\wedge\sim{r}\vee\sim{p}\wedge{r} \\ $$$$={p}\vee{q}\vee{r}\wedge\left(\sim{p}\vee\sim{r}\right) \\ $$