Sr Examen
Expresión ¬¬(¬a⇒b)⇒c⇔¬a⇒b∨c
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Solución
Ha introducido
[src]
¬(((¬a)⇒(b∨c))⇔((¬((¬a)⇒b))⇒c))
$$\left(\neg a \Rightarrow \left(b \vee c\right)\right) \not\equiv \left(\neg a \not\Rightarrow b \Rightarrow c\right)$$
Solución detallada
$$\neg a \Rightarrow \left(b \vee c\right) = a \vee b \vee c$$
$$\neg a \Rightarrow b = a \vee b$$
$$\neg a \not\Rightarrow b = \neg a \wedge \neg b$$
$$\neg a \not\Rightarrow b \Rightarrow c = a \vee b \vee c$$
$$\left(\neg a \Rightarrow \left(b \vee c\right)\right) ⇔ \left(\neg a \not\Rightarrow b \Rightarrow c\right) = 1$$
$$\left(\neg a \Rightarrow \left(b \vee c\right)\right) \not\equiv \left(\neg a \not\Rightarrow b \Rightarrow c\right) = \text{False}$$
$$\neg a \Rightarrow b = a \vee b$$
$$\neg a \not\Rightarrow b = \neg a \wedge \neg b$$
$$\neg a \not\Rightarrow b \Rightarrow c = a \vee b \vee c$$
$$\left(\neg a \Rightarrow \left(b \vee c\right)\right) ⇔ \left(\neg a \not\Rightarrow b \Rightarrow c\right) = 1$$
$$\left(\neg a \Rightarrow \left(b \vee c\right)\right) \not\equiv \left(\neg a \not\Rightarrow b \Rightarrow c\right) = \text{False}$$
Tabla de verdad
+---+---+---+--------+ | a | b | c | result | +===+===+===+========+ | 0 | 0 | 0 | 0 | +---+---+---+--------+ | 0 | 0 | 1 | 0 | +---+---+---+--------+ | 0 | 1 | 0 | 0 | +---+---+---+--------+ | 0 | 1 | 1 | 0 | +---+---+---+--------+ | 1 | 0 | 0 | 0 | +---+---+---+--------+ | 1 | 0 | 1 | 0 | +---+---+---+--------+ | 1 | 1 | 0 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+