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Peter

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1^st lesson free

Maths
Trigonometry
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Science
Calculation

Hat (𝑝 ∧ 𝑞)≡T, must be that 𝑝≡T and 𝑞≡T, hence [(𝑝 ∧ 𝑞) → 𝑟]≡ 𝑟, and [[(𝑝 ∧ 𝑞) → 𝑟] → 𝑠]≡ 𝑟 → F≡¬𝑟. Secondly, [(𝑝 → 𝑟) → 𝑠] from t

Name: Peter
Address: Auckland, PH
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Maths
Trigonometry
Arithmetic
Science
Calculation

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About Peter

Therefore proving that [[(𝑝 ∧ 𝑞) → 𝑟] → 𝑠] → [(𝑝 → 𝑟) → 𝑠] ≡ T. Q.E.DTherefore proving that [[(𝑝 ∧ 𝑞) → 𝑟] → 𝑠] → [(𝑝 → 𝑟) → 𝑠] ≡ T. Q.E.DTherefore proving that [[(𝑝 ∧ 𝑞) → 𝑟] → 𝑠] → [(𝑝 → 𝑟) → 𝑠] ≡ T. Q.E.D

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herefore 𝑝 ∨ (𝑞 ∨ 𝑟)≡F. So for the left side, since no matter if it is 𝑝≡F or
(𝑞 ∨ 𝑟)≡F, one out of the two possibility being false is certain, so necessarily 𝑝 ∧ (𝑞 ∨ 𝑟)≡F, and since F F≡T, [𝑝 ∧ (𝑞 ∨ 𝑟)] [(𝑝 ∧ 𝑞) ∨ (𝑝 ∧ 𝑟)] is true.

And consider the second case where (𝑝 ∧ 𝑞) ∨ (𝑝 ∧ 𝑟)≡T, for it to be held true, necessarily (𝑝 ∧ 𝑞)≡T or (𝑝 ∧ 𝑟)≡T. Hence either 𝑝 ∧ (𝑞 ∧ 𝑟) ≡T or 𝑝 ∧ (𝑞 ∧ 𝑟) ≡F. Thus for 𝑝 ∧ (𝑞 ∨ 𝑟), its truth value must be T.

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Hat (𝑝 ∧ 𝑞)≡T, must be that 𝑝≡T and 𝑞≡T, hence [(𝑝 ∧ 𝑞) → 𝑟]≡ 𝑟, and [[(𝑝 ∧ 𝑞) → 𝑟] → 𝑠]≡ 𝑟 → F≡¬𝑟. Secondly, [(𝑝 → 𝑟) → 𝑠] from t

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Hat (𝑝 ∧ 𝑞)≡T, must be that 𝑝≡T and 𝑞≡T, hence [(𝑝 ∧ 𝑞) → 𝑟]≡ 𝑟, and [[(𝑝 ∧ 𝑞) → 𝑟] → 𝑠]≡ 𝑟 → F≡¬𝑟. Secondly, [(𝑝 → 𝑟) → 𝑠] from t

Lesson location

About Peter

About the lesson

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Rate

Pack rates

free lessons

online

Other tutors in Maths

Tristan

Lenard

Sophia

Earl Jhon

Kacey(Kia)

Jenny Mae

Edward

John Paul

Princess Karen

Mary Jane

Christian Mhel

Whey

Ellie

Kenny Jay

MARVIN SAGRADO

Miraflor

RENDIE C

Danniel Raphael

Glemhar

Cheska

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