Sr Examen
Expresión BA⇒(A⇒B)⇔CB
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Solución
Ha introducido
[src]
(b∧c)⇔((a∧b)⇒(a⇒b))
$$\left(b \wedge c\right) ⇔ \left(\left(a \wedge b\right) \Rightarrow \left(a \Rightarrow b\right)\right)$$
Solución detallada
$$a \Rightarrow b = b \vee \neg a$$
$$\left(a \wedge b\right) \Rightarrow \left(a \Rightarrow b\right) = 1$$
$$\left(b \wedge c\right) ⇔ \left(\left(a \wedge b\right) \Rightarrow \left(a \Rightarrow b\right)\right) = b \wedge c$$
$$\left(a \wedge b\right) \Rightarrow \left(a \Rightarrow b\right) = 1$$
$$\left(b \wedge c\right) ⇔ \left(\left(a \wedge b\right) \Rightarrow \left(a \Rightarrow b\right)\right) = b \wedge c$$
Tabla de verdad
+---+---+---+--------+ | a | b | c | result | +===+===+===+========+ | 0 | 0 | 0 | 0 | +---+---+---+--------+ | 0 | 0 | 1 | 0 | +---+---+---+--------+ | 0 | 1 | 0 | 0 | +---+---+---+--------+ | 0 | 1 | 1 | 1 | +---+---+---+--------+ | 1 | 0 | 0 | 0 | +---+---+---+--------+ | 1 | 0 | 1 | 0 | +---+---+---+--------+ | 1 | 1 | 0 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+