Sr Examen
Expresión (¬(b∧c)→(a∨c))→¬(¬(a∨b)→(c→¬b))
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
((¬(b∧c))⇒(a∨c))⇒(¬((¬(a∨b))⇒(c⇒(¬b))))
$$\left(\neg \left(b \wedge c\right) \Rightarrow \left(a \vee c\right)\right) \Rightarrow \neg \left(a \vee b\right) \not\Rightarrow \left(c \Rightarrow \neg b\right)$$
Solución detallada
$$\neg \left(b \wedge c\right) = \neg b \vee \neg c$$
$$\neg \left(b \wedge c\right) \Rightarrow \left(a \vee c\right) = a \vee c$$
$$\neg \left(a \vee b\right) = \neg a \wedge \neg b$$
$$c \Rightarrow \neg b = \neg b \vee \neg c$$
$$\neg \left(a \vee b\right) \Rightarrow \left(c \Rightarrow \neg b\right) = 1$$
$$\neg \left(a \vee b\right) \not\Rightarrow \left(c \Rightarrow \neg b\right) = \text{False}$$
$$\left(\neg \left(b \wedge c\right) \Rightarrow \left(a \vee c\right)\right) \Rightarrow \neg \left(a \vee b\right) \not\Rightarrow \left(c \Rightarrow \neg b\right) = \neg a \wedge \neg c$$
$$\neg \left(b \wedge c\right) \Rightarrow \left(a \vee c\right) = a \vee c$$
$$\neg \left(a \vee b\right) = \neg a \wedge \neg b$$
$$c \Rightarrow \neg b = \neg b \vee \neg c$$
$$\neg \left(a \vee b\right) \Rightarrow \left(c \Rightarrow \neg b\right) = 1$$
$$\neg \left(a \vee b\right) \not\Rightarrow \left(c \Rightarrow \neg b\right) = \text{False}$$
$$\left(\neg \left(b \wedge c\right) \Rightarrow \left(a \vee c\right)\right) \Rightarrow \neg \left(a \vee b\right) \not\Rightarrow \left(c \Rightarrow \neg b\right) = \neg a \wedge \neg c$$
Tabla de verdad
+---+---+---+--------+ | a | b | c | result | +===+===+===+========+ | 0 | 0 | 0 | 1 | +---+---+---+--------+ | 0 | 0 | 1 | 0 | +---+---+---+--------+ | 0 | 1 | 0 | 1 | +---+---+---+--------+ | 0 | 1 | 1 | 0 | +---+---+---+--------+ | 1 | 0 | 0 | 0 | +---+---+---+--------+ | 1 | 0 | 1 | 0 | +---+---+---+--------+ | 1 | 1 | 0 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+