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Question Number 191735 by Spillover last updated on 29/Apr/23

Verify that   ┐(p→q)→(p∧^┐ q) is tautology using laws of  algebra

$${Verify}\:{that} \\ $$$$\:\urcorner\left({p}\rightarrow{q}\right)\rightarrow\left({p}\wedge^{\urcorner} {q}\right)\:{is}\:{tautology}\:{using}\:{laws}\:{of} \\ $$$${algebra} \\ $$

Answered by manxsol last updated on 30/Apr/23

∼(∼p q)→(p ∼q)  (∼p q) (p ∼q)  (∼p q) p} (∼p q ∼q)  {V  q} {V  q)  V V=V tautologia  p→q=∼p q  distributiva  p (q r)=p q) (p r)

$$\sim\left(\sim{p} {q}\right)\rightarrow\left({p} \sim{q}\right) \\ $$$$\left(\sim{p} {q}\right) \left({p} \sim{q}\right) \\ $$$$\left.\left(\sim{p} {q}\right) {p}\right\} \left(\sim{p} {q} \sim{q}\right) \\ $$$$\left\{\boldsymbol{{V}}\: {q}\right\} \left\{\boldsymbol{{V}}\: {q}\right) \\ $$$${V} {V}={V}\:{tautologia} \\ $$$${p}\rightarrow{q}=\sim{p} {q} \\ $$$${distributiva} \\ $$$$\left.{p} \left({q} {r}\right)={p} {q}\right) \left({p} {r}\right) \\ $$

Answered by aba last updated on 30/Apr/23

 determinant ((p,q,(⌉(p→q)),(p∧⌉q),(⌉(p→q)→(p∧⌉q))),(0,0,0,0,1),(0,1,0,0,1),(1,0,1,1,1),(1,1,0,0,1))

$$\begin{array}{|c|c|c|c|c|}{\mathrm{p}}&\hline{\mathrm{q}}&\hline{\rceil\left(\mathrm{p}\rightarrow\mathrm{q}\right)}&\hline{\mathrm{p}\wedge\rceil\mathrm{q}}&\hline{\rceil\left(\mathrm{p}\rightarrow\mathrm{q}\right)\rightarrow\left(\mathrm{p}\wedge\rceil\mathrm{q}\right)}\\{\mathrm{0}}&\hline{\mathrm{0}}&\hline{\mathrm{0}}&\hline{\mathrm{0}}&\hline{\mathrm{1}}\\{\mathrm{0}}&\hline{\mathrm{1}}&\hline{\mathrm{0}}&\hline{\mathrm{0}}&\hline{\mathrm{1}}\\{\mathrm{1}}&\hline{\mathrm{0}}&\hline{\mathrm{1}}&\hline{\mathrm{1}}&\hline{\mathrm{1}}\\{\mathrm{1}}&\hline{\mathrm{1}}&\hline{\mathrm{0}}&\hline{\mathrm{0}}&\hline{\mathrm{1}}\\\hline\end{array} \\ $$

Commented by Spillover last updated on 30/Apr/23

read the question carefull.the question says  use the laws of algebra.Thanks for your solution

$${read}\:{the}\:{question}\:{carefull}.{the}\:{question}\:{says} \\ $$$${use}\:{the}\:{laws}\:{of}\:{algebra}.{Thanks}\:{for}\:{your}\:{solution} \\ $$

Commented by aba last updated on 30/Apr/23

i know

$$\mathrm{i}\:\mathrm{know} \\ $$

Answered by Spillover last updated on 30/Apr/23

∼(p→q)→(p∧∼q)  ∼[(∼p∨q)∨∼(p∧∼q)    p∧∼q∨∼p∨q  p∨∼p∼q∨q  T∨T  T

$$\sim\left({p}\rightarrow{q}\right)\rightarrow\left({p}\wedge\sim{q}\right) \\ $$$$\sim\left[\left(\sim{p}\vee{q}\right)\vee\sim\left({p}\wedge\sim{q}\right)\:\:\right. \\ $$$${p}\wedge\sim{q}\vee\sim{p}\vee{q} \\ $$$${p}\vee\sim{p}\sim{q}\vee{q} \\ $$$${T}\vee{T} \\ $$$${T} \\ $$

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