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

Mathematics
Trigonometry
Arithmetic
Science
Calculus

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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.

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 rates

online

Other tutors in Mathematics

Sofia

Philip

Reza

Jose

Josiah

Lyle

Drew

Davayne

Joe

Fadil

Boris

Natalie

Seren

Bryan

Matias Ignacio

Yannik

Jody

Sevan

Haider

Cynthia

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