Peter

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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

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

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

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Gcse/igcse and a level examiner with uk boards,11+, 13+ and 16+ entrances, 100% grade 7-9, a/a*, excellent results, 20yrs experience leading schools and working in schools/sixth form colleges

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