Peter - Maths tutor - Auckland
1st lesson belanja
Peter - Maths tutor - Auckland

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Peter

  • Rate RM117
  • Response 1h
Peter - Maths tutor - Auckland

RM117/h

1st lesson belanja

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1st lesson belanja

1st lesson belanja

  • Maths
  • Trigonometry
  • Arithmetic
  • School science

Hat (𝑝 ∧ π‘ž)≑T, must be that 𝑝≑T and π‘žβ‰‘T, hence [(𝑝 ∧ π‘ž) β†’ π‘Ÿ]≑ π‘Ÿ, and [[(𝑝 ∧ π‘ž) β†’ π‘Ÿ] β†’ 𝑠]≑ π‘Ÿ β†’ Fβ‰‘Β¬π‘Ÿ. Secondly, [(𝑝 β†’ π‘Ÿ) β†’ 𝑠] from t

  • Maths
  • Trigonometry
  • Arithmetic
  • School science

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Therefore proving that [[(𝑝 ∧ π‘ž) β†’ π‘Ÿ] β†’ 𝑠] β†’ [(𝑝 β†’ π‘Ÿ) β†’ 𝑠] ≑ T. Q.E.DTherefore proving that [[(𝑝 ∧ π‘ž) β†’ π‘Ÿ] β†’ 𝑠] β†’ [(𝑝 β†’ π‘Ÿ) β†’ 𝑠] ≑ T. Q.E.DTherefore proving that [[(𝑝 ∧ π‘ž) β†’ π‘Ÿ] β†’ 𝑠] β†’ [(𝑝 β†’ π‘Ÿ) β†’ 𝑠] ≑ T. Q.E.D

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  • SPM
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English

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

And consider the second case where (𝑝 ∧ π‘ž) ∨ (𝑝 ∧ π‘Ÿ)≑T, for it to be held true, necessarily (𝑝 ∧ π‘ž)≑T or (𝑝 ∧ π‘Ÿ)≑T. Hence either 𝑝 ∧ (π‘ž ∧ π‘Ÿ) ≑T or 𝑝 ∧ (π‘ž ∧ π‘Ÿ) ≑F. Thus for 𝑝 ∧ (π‘ž ∨ π‘Ÿ), its truth value must be T.

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  • RM117

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  • 5h: RM585
  • 10h: RM1170

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  • RM117/h

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