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Expresión avc>a<=>(c&(¬a))<=>c
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Solución
Ha introducido
[src]
c⇔(c∧(¬a))⇔((a∨c)⇒a)
$$c ⇔ \left(c \wedge \neg a\right) ⇔ \left(\left(a \vee c\right) \Rightarrow a\right)$$
Solución detallada
$$\left(a \vee c\right) \Rightarrow a = a \vee \neg c$$
$$c ⇔ \left(c \wedge \neg a\right) ⇔ \left(\left(a \vee c\right) \Rightarrow a\right) = \text{False}$$
$$c ⇔ \left(c \wedge \neg a\right) ⇔ \left(\left(a \vee c\right) \Rightarrow a\right) = \text{False}$$
Tabla de verdad
+---+---+--------+ | a | c | result | +===+===+========+ | 0 | 0 | 0 | +---+---+--------+ | 0 | 1 | 0 | +---+---+--------+ | 1 | 0 | 0 | +---+---+--------+ | 1 | 1 | 0 | +---+---+--------+