Let Δ,{,} be such that pq((pΔq)r) is a tautology. Then (pq)Δr is logically equivalent to

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mrow> <mo>(</mo> <mrow> <mi>p</mi> <mi>Δ</mi> <mi>r</mi> </mrow> <mo>)</mo> </mrow> <mo>∨</mo> <mi>q</mi> </mrow> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mrow> <mo>(</mo> <mrow> <mi>p</mi> <mi>Δ</mi> <mi>r</mi> </mrow> <mo>)</mo> </mrow> <mo>∧</mo> <mi>q</mi> </mrow> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mrow> <mo>(</mo> <mrow> <mi>p</mi> <mo>∧</mo> <mi>r</mi> </mrow> <mo>)</mo> </mrow> <mi>Δ</mi> <mi>q</mi> </mrow> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mrow> <mo>(</mo> <mrow> <mi>p</mi> <mo>∇</mo> <mi>r</mi> </mrow> <mo>)</mo> </mrow> <mo>∧</mo> <mi>q</mi> </mrow> </math> </span></p>
2 Views|Posted 7 months ago
Asked by Shiksha User
1 Answer
P
7 months ago
Correct Option - 1
Detailed Solution:

Case – I Δ

(pq) ( (pq)r)

it can be false if r is false,

so not a tautology

Case – II If Δ

(pq)? ( (pq)r) tautology

then  (pq)r (pΔr)q

Case – III If Δv,

then  (pq) { (pq)r}

Not a tautology

Case – IV If Δ,

(pq) { (pq)r}

Not a tautology

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