Why is this statement of logical equivalence true? f ∨ g ∧ h ≡ (f ∨ g) ∧ (f ∨ h)
Why is this statement of logical equivalence true?
f ∨ g ∧ h ≡ (f ∨ g) ∧ (f ∨ h)
I made the truth tables below, and this statement doesn´t hold--it isn't a logical equivalence. However, according to the exercise it is.
f ∨ g ∧ h:
(f ∨ g) ∧ (f ∨ h):
Have a close look at the way the equivalence is denoted:
no parenthesis on the left side, two on the right specifying the disjunctions to be evaluated first.
If this was the conventional logical operator precedence, both pairs of parentheses were dispensable.
If left to right evaluation was conventional, the first pair was redundant: the convention actually is conjunction before disjunction (and the equivalence is denoted correctly as well as irredundant).
So, the f ∨ g column in your first table is immaterial, if correct.
The f ∨ g ∧ h column is inconsistent: the label with explicit parentheses would need to be f ∨ (g ∧ h), the values tabulated are for (f ∨ g) ∧ h.
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