Peter

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Maths
Trigonometry
Arithmetic
School science
Calculus

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

Maths
Trigonometry
Arithmetic
School science
Calculus

Lesson location

- At Peter's house: Auckland
- online
- at home or in a public space : will travel up to 20km from Auckland

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

About the lesson

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.

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

Rates

Rate

Pack prices

online

Similar Maths tutors in Auckland

Kajana

Osama

Evan

Ho Ming (Hoffman)

Doris

Steven

Jessica Ji

Ahmed

Sherif

Grace

Lakmini

Nandana

Denisse

Ethan

Riya

Faheem

Kairav

Soshi

Xiaolin

Melody

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