Sr Examen
Expresión A→(B∨C)≡(A→B)∨(A→C)
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
(a⇒(b∨c))⇔((a⇒b)∨(a⇒c))
$$\left(a \Rightarrow \left(b \vee c\right)\right) ⇔ \left(\left(a \Rightarrow b\right) \vee \left(a \Rightarrow c\right)\right)$$
Solución detallada
$$a \Rightarrow \left(b \vee c\right) = b \vee c \vee \neg a$$
$$a \Rightarrow b = b \vee \neg a$$
$$a \Rightarrow c = c \vee \neg a$$
$$\left(a \Rightarrow b\right) \vee \left(a \Rightarrow c\right) = b \vee c \vee \neg a$$
$$\left(a \Rightarrow \left(b \vee c\right)\right) ⇔ \left(\left(a \Rightarrow b\right) \vee \left(a \Rightarrow c\right)\right) = 1$$
$$a \Rightarrow b = b \vee \neg a$$
$$a \Rightarrow c = c \vee \neg a$$
$$\left(a \Rightarrow b\right) \vee \left(a \Rightarrow c\right) = b \vee c \vee \neg a$$
$$\left(a \Rightarrow \left(b \vee c\right)\right) ⇔ \left(\left(a \Rightarrow b\right) \vee \left(a \Rightarrow c\right)\right) = 1$$
Tabla de verdad
+---+---+---+--------+ | a | b | c | result | +===+===+===+========+ | 0 | 0 | 0 | 1 | +---+---+---+--------+ | 0 | 0 | 1 | 1 | +---+---+---+--------+ | 0 | 1 | 0 | 1 | +---+---+---+--------+ | 0 | 1 | 1 | 1 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+