Задачи по дискретной математике для контрольных и самостоятельных работ. Булевы функции. Васильев А.В - 6 стр.

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4. Проверить эквивалентность формул Φ и Ψ, построив таблицы.
4.1. Φ = x
1
(x
2
∨ x
3
), Ψ = x
1
x
2
∨ x
1
x
3
4.2. Φ = x
1
(x
2
⊕ x
3
), Ψ = x
1
x
2
⊕ x
1
x
3
4.3. Φ = x
1
(x
2
∼ x
3
), Ψ = x
1
x
2
∼ x
1
x
3
4.4. Φ = x
1
(x
2
→ x
3
), Ψ = x
1
x
2
→ x
1
x
3
4.5. Φ = x
1
(x
2
| x
3
), Ψ = (x
1
x
2
) | (x
1
x
3
)
4.6. Φ = x
1
(x
2
↓ x
3
), Ψ = (x
1
x
2
) ↓ (x
1
x
3
)
4.7. Φ = x
1
∨ (x
2
x
3
), Ψ = (x
1
∨ x
2
)(x
1
∨ x
3
)
4.8. Φ = x
1
∨ (x
2
⊕ x
3
), Ψ = (x
1
∨ x
2
) ⊕ (x
1
∨ x
3
)
4.9. Φ = x
1
∨ (x
2
∼ x
3
), Ψ = (x
1
∨ x
2
) ∼ (x
1
∨ x
3
)
4.10. Φ = x
1
∨ (x
2
→ x
3
), Ψ = (x
1
∨ x
2
) → (x
1
∨ x
3
)
4.11. Φ = x
1
∨ (x
2
| x
3
), Ψ = (x
1
∨ x
2
) | (x
1
∨ x
3
)
4.12. Φ = x
1
∨ (x
2
↓ x
3
), Ψ = (x
1
∨ x
2
) ↓ (x
1
∨ x
3
)
4.13. Φ = x
1
⊕ (x
2
x
3
), Ψ = (x
1
⊕ x
2
)(x
1
⊕ x
3
)
4.14. Φ = x
1
⊕ (x
2
∨ x
3
), Ψ = (x
1
⊕ x
2
) ∨ (x
1
⊕ x
3
)
4.15. Φ = x
1
⊕ (x
2
∼ x
3
), Ψ = (x
1
⊕ x
2
) ∼ (x
1
⊕ x
3
)
4.16. Φ = x
1
⊕ (x
2
→ x
3
), Ψ = (x
1
⊕ x
2
) → (x
1
⊕ x
3
)
4.17. Φ = x
1
⊕ (x
2
| x
3
), Ψ = (x
1
⊕ x
2
) | (x
1
⊕ x
3
)
4.18. Φ = x
1
⊕ (x
2
↓ x
3
), Ψ = (x
1
⊕ x
2
) ↓ (x
1
⊕ x
3
)
4.19. Φ = x
1
| (x
2
x
3
), Ψ = (x
1
| x
2
) ∨ (x
1
| x
3
)
4.20. Φ = x
1
| (x
2
∨ x
3
), Ψ = (x
1
| x
2
)(x
1
| x
3
)
4.21. Φ = x
1
| (x
2
⊕ x
3
), Ψ = (x
1
| x
2
) ⊕ (x
1
| x
3
)
4.22. Φ = x
1
| (x
2
∼ x
3
), Ψ = (x
1
| x
2
) ∼ (x
1
| x
3
)
4.23. Φ = x
1
| (x
2
→ x
3
), Ψ = (x
1
| x
2
) → (x
1
| x
3
)
4.24. Φ = x
1
| (x
2
↓ x
3
), Ψ = (x
1
| x
2
) ↓ (x
1
| x
3
)
4.25. Φ = x
1
↓ (x
2
x
3
), Ψ = (x
1
↓ x
2
) ∨ (x
1
↓ x
3
)
4.26. Φ = x
1
↓ (x
2
∨ x
3
), Ψ = (x
1
↓ x
2
)(x
1
↓ x
3
)
4.27. Φ = x
1
↓ (x
2
⊕ x
3
), Ψ = (x
1
↓ x
2
) ⊕ (x
1
↓ x
3
)
4.28. Φ = x
1
↓ (x
2
∼ x
3
), Ψ = (x
1
↓ x
2
) ∼ (x
1
↓ x
3
)
4.29. Φ = x
1
↓ (x
2
→ x
3
), Ψ = (x
1
↓ x
2
) → (x
1
↓ x
3
)
4.30. Φ = x
1
↓ (x
2
| x
3
), Ψ = (x
1
↓ x
2
) | (x
1
↓ x
3
)
6
4. Проверить эквивалентность формул Φ и Ψ, построив таблицы.
 4.1. Φ = x1 (x2 ∨ x3 ), Ψ = x1 x2 ∨ x1 x3
 4.2. Φ = x1 (x2 ⊕ x3 ), Ψ = x1 x2 ⊕ x1 x3
 4.3. Φ = x1 (x2 ∼ x3 ), Ψ = x1 x2 ∼ x1 x3
 4.4. Φ = x1 (x2 → x3 ), Ψ = x1 x2 → x1 x3
 4.5. Φ = x1 (x2 | x3 ), Ψ = (x1 x2 ) | (x1 x3 )
 4.6. Φ = x1 (x2 ↓ x3 ), Ψ = (x1 x2 ) ↓ (x1 x3 )
 4.7. Φ = x1 ∨ (x2 x3 ), Ψ = (x1 ∨ x2 )(x1 ∨ x3 )
 4.8. Φ = x1 ∨ (x2 ⊕ x3 ), Ψ = (x1 ∨ x2 ) ⊕ (x1 ∨ x3 )
 4.9. Φ = x1 ∨ (x2 ∼ x3 ), Ψ = (x1 ∨ x2 ) ∼ (x1 ∨ x3 )
4.10. Φ = x1 ∨ (x2 → x3 ), Ψ = (x1 ∨ x2 ) → (x1 ∨ x3 )
4.11. Φ = x1 ∨ (x2 | x3 ), Ψ = (x1 ∨ x2 ) | (x1 ∨ x3 )
4.12. Φ = x1 ∨ (x2 ↓ x3 ), Ψ = (x1 ∨ x2 ) ↓ (x1 ∨ x3 )
4.13. Φ = x1 ⊕ (x2 x3 ), Ψ = (x1 ⊕ x2 )(x1 ⊕ x3 )
4.14. Φ = x1 ⊕ (x2 ∨ x3 ), Ψ = (x1 ⊕ x2 ) ∨ (x1 ⊕ x3 )
4.15. Φ = x1 ⊕ (x2 ∼ x3 ), Ψ = (x1 ⊕ x2 ) ∼ (x1 ⊕ x3 )
4.16. Φ = x1 ⊕ (x2 → x3 ), Ψ = (x1 ⊕ x2 ) → (x1 ⊕ x3 )
4.17. Φ = x1 ⊕ (x2 | x3 ), Ψ = (x1 ⊕ x2 ) | (x1 ⊕ x3 )
4.18. Φ = x1 ⊕ (x2 ↓ x3 ), Ψ = (x1 ⊕ x2 ) ↓ (x1 ⊕ x3 )
4.19. Φ = x1 | (x2 x3 ), Ψ = (x1 | x2 ) ∨ (x1 | x3 )
4.20. Φ = x1 | (x2 ∨ x3 ), Ψ = (x1 | x2 )(x1 | x3 )
4.21. Φ = x1 | (x2 ⊕ x3 ), Ψ = (x1 | x2 ) ⊕ (x1 | x3 )
4.22. Φ = x1 | (x2 ∼ x3 ), Ψ = (x1 | x2 ) ∼ (x1 | x3 )
4.23. Φ = x1 | (x2 → x3 ), Ψ = (x1 | x2 ) → (x1 | x3 )
4.24. Φ = x1 | (x2 ↓ x3 ), Ψ = (x1 | x2 ) ↓ (x1 | x3 )
4.25. Φ = x1 ↓ (x2 x3 ), Ψ = (x1 ↓ x2 ) ∨ (x1 ↓ x3 )
4.26. Φ = x1 ↓ (x2 ∨ x3 ), Ψ = (x1 ↓ x2 )(x1 ↓ x3 )
4.27. Φ = x1 ↓ (x2 ⊕ x3 ), Ψ = (x1 ↓ x2 ) ⊕ (x1 ↓ x3 )
4.28. Φ = x1 ↓ (x2 ∼ x3 ), Ψ = (x1 ↓ x2 ) ∼ (x1 ↓ x3 )
4.29. Φ = x1 ↓ (x2 → x3 ), Ψ = (x1 ↓ x2 ) → (x1 ↓ x3 )
4.30. Φ = x1 ↓ (x2 | x3 ), Ψ = (x1 ↓ x2 ) | (x1 ↓ x3 )



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