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