Sr Examen
Expresión ¬(A∧B⇒A)∨A∧(B∨C)
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Solución
Ha introducido
[src]
(a∧(b∨c))∨(¬((a∧b)⇒a))
$$\left(a \wedge \left(b \vee c\right)\right) \vee \left(a \wedge b\right) \not\Rightarrow a$$
Solución detallada
$$\left(a \wedge b\right) \Rightarrow a = 1$$
$$\left(a \wedge b\right) \not\Rightarrow a = \text{False}$$
$$\left(a \wedge \left(b \vee c\right)\right) \vee \left(a \wedge b\right) \not\Rightarrow a = a \wedge \left(b \vee c\right)$$
$$\left(a \wedge b\right) \not\Rightarrow a = \text{False}$$
$$\left(a \wedge \left(b \vee c\right)\right) \vee \left(a \wedge b\right) \not\Rightarrow a = a \wedge \left(b \vee c\right)$$
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 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+