Question #267239

Let D = {-5, -3, -1, 1, 3, 5}. Write the following statements using only negations, conjunctions and


disjunctions:


a) âˆƒđ‘„đ‘ƒ(đ‘„)


b) âˆ€đ‘„đ‘ƒ(đ‘„)


c) âˆ€đ‘„((đ‘„ ≠ 1) → 𝑃(đ‘„))


d) âˆƒđ‘„((đ‘„ ≄ 0) ∧ 𝑃(đ‘„))


e) âˆƒđ‘„(ïżąđ‘ƒ(đ‘„)) ∧ âˆ€đ‘„((đ‘„ < 0) → 𝑃(đ‘„))

1
Expert's answer
2021-11-17T15:08:50-0500

a) âˆƒxP(x)=P(−5)√P(−3)√P(−1)√P(1)√P(3)√P(5)∃xP(x)=P(-5)\lor P(-3)\lor P(-1)\lor P(1)\lor P(3)\lor P(5)


b) âˆ€xP(x)=P(−5)∧P(−3)∧P(−1)∧P(1)∧P(3)∧P(5)∀xP(x)=P(-5)\land P(-3)\land P(-1)\land P(1)\land P(3)\land P(5)


c)

âˆ€đ‘„((đ‘„â‰ 1)→𝑃(đ‘„))=âˆ€đ‘„(ÂŹ(x≠1)√P(x))=(x=1)√P(1)âˆ€đ‘„((đ‘„ ≠ 1) → 𝑃(đ‘„))=âˆ€đ‘„(\neg(x\neq 1)\lor P(x))=(x=1)\lor P(1)

d)

âˆƒđ‘„((đ‘„â‰„0)∧𝑃(đ‘„))=((x=1)∧P(1))√((x=)∧P(3))√((x=5)∧P(5))âˆƒđ‘„((đ‘„ ≄ 0) ∧ 𝑃(đ‘„))=((x=1)\land P(1))\lor ((x=)\land P(3))\lor ((x=5)\land P(5))


e)

âˆƒđ‘„(ïżąđ‘ƒ(đ‘„))âˆ§âˆ€đ‘„((đ‘„<0)→𝑃(đ‘„))=âˆƒđ‘„(ïżąđ‘ƒ(đ‘„))âˆ§âˆ€đ‘„(ÂŹ(đ‘„<0)∹𝑃(đ‘„))=âˆƒđ‘„(ïżąđ‘ƒ(đ‘„)) ∧ âˆ€đ‘„((đ‘„ < 0) → 𝑃(đ‘„)) =âˆƒđ‘„(ïżąđ‘ƒ(đ‘„)) ∧ âˆ€đ‘„(\neg(đ‘„ < 0) \lor 𝑃(đ‘„))=

=âˆƒđ‘„(ïżąđ‘ƒ(đ‘„))âˆ§âˆ€đ‘„((đ‘„>0)∹𝑃(đ‘„))==âˆƒđ‘„(ïżąđ‘ƒ(đ‘„)) ∧ âˆ€đ‘„((đ‘„ > 0) \lor 𝑃(đ‘„))=

=âˆƒđ‘„(ïżąđ‘ƒ(đ‘„))∧(((đ‘„=1)∹𝑃(1))∧((đ‘„=3)∹𝑃(3))∧((đ‘„=5)∹𝑃(5)))==âˆƒđ‘„(ïżąđ‘ƒ(đ‘„)) ∧ (((đ‘„ =1) \lor 𝑃(1))\land ((đ‘„ =3) \lor 𝑃(3))\land ((đ‘„ =5) \lor 𝑃(5)))=


=(ïżąđ‘ƒ(−5))∧(((đ‘„=1)∹𝑃(1))∧((đ‘„=3)∹𝑃(3))∧((đ‘„=5)∹𝑃(5)))√=(ïżąđ‘ƒ(-5)) ∧ (((đ‘„ =1) \lor 𝑃(1))\land ((đ‘„ =3) \lor 𝑃(3))\land ((đ‘„ =5) \lor 𝑃(5)))\lor

√(ïżąđ‘ƒ(−3))∧(((đ‘„=1)∹𝑃(đ‘„))∧((đ‘„=3)∹𝑃(3))∧((đ‘„=5)∹𝑃(5)))√\lor (ïżąđ‘ƒ(-3)) ∧ (((đ‘„ =1) \lor 𝑃(đ‘„))\land ((đ‘„ =3) \lor 𝑃(3))\land ((đ‘„ =5) \lor 𝑃(5)))\lor

√(ïżąđ‘ƒ(−1))∧(((đ‘„=1)∹𝑃(đ‘„))∧((đ‘„=3)∹𝑃(3))∧((đ‘„=5)∹𝑃(5)))√\lor (ïżąđ‘ƒ(-1)) ∧ (((đ‘„ =1) \lor 𝑃(đ‘„))\land ((đ‘„ =3) \lor 𝑃(3))\land ((đ‘„ =5) \lor 𝑃(5)))\lor

√(ïżąđ‘ƒ(1))∧(((đ‘„=1)∹𝑃(đ‘„))∧((đ‘„=3)∹𝑃(3))∧((đ‘„=5)∹𝑃(5)))√\lor (ïżąđ‘ƒ(1)) ∧ (((đ‘„ =1) \lor 𝑃(đ‘„))\land ((đ‘„ =3) \lor 𝑃(3))\land ((đ‘„ =5) \lor 𝑃(5)))\lor

√(ïżąđ‘ƒ(3))∧(((đ‘„=1)∹𝑃(đ‘„))∧((đ‘„=3)∹𝑃(3))∧((đ‘„=5)∹𝑃(5)))√\lor (ïżąđ‘ƒ(3)) ∧ (((đ‘„ =1) \lor 𝑃(đ‘„))\land ((đ‘„ =3) \lor 𝑃(3))\land ((đ‘„ =5) \lor 𝑃(5)))\lor

√(ïżąđ‘ƒ(5))∧(((đ‘„=1)∹𝑃(đ‘„))∧((đ‘„=3)∹𝑃(3))∧((đ‘„=5)∹𝑃(5)))\lor (ïżąđ‘ƒ(5)) ∧ (((đ‘„ =1) \lor 𝑃(đ‘„))\land ((đ‘„ =3) \lor 𝑃(3))\land ((đ‘„ =5) \lor 𝑃(5)))

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