Sr Examen
Expresión (A↔B)→((A∧C)→(A→B∨C))
El profesor se sorprenderá mucho al ver tu solución correcta😉
⌨
Solución
Ha introducido
[src]
(a⇔b)⇒((a∧c)⇒(a⇒(b∨c)))
$$\left(a ⇔ b\right) \Rightarrow \left(\left(a \wedge c\right) \Rightarrow \left(a \Rightarrow \left(b \vee c\right)\right)\right)$$
Solución detallada
$$a ⇔ b = \left(a \wedge b\right) \vee \left(\neg a \wedge \neg b\right)$$
$$a \Rightarrow \left(b \vee c\right) = b \vee c \vee \neg a$$
$$\left(a \wedge c\right) \Rightarrow \left(a \Rightarrow \left(b \vee c\right)\right) = 1$$
$$\left(a ⇔ b\right) \Rightarrow \left(\left(a \wedge c\right) \Rightarrow \left(a \Rightarrow \left(b \vee c\right)\right)\right) = 1$$
$$a \Rightarrow \left(b \vee c\right) = b \vee c \vee \neg a$$
$$\left(a \wedge c\right) \Rightarrow \left(a \Rightarrow \left(b \vee c\right)\right) = 1$$
$$\left(a ⇔ b\right) \Rightarrow \left(\left(a \wedge c\right) \Rightarrow \left(a \Rightarrow \left(b \vee c\right)\right)\right) = 1$$
Tabla de verdad
+---+---+---+--------+ | a | b | c | result | +===+===+===+========+ | 0 | 0 | 0 | 1 | +---+---+---+--------+ | 0 | 0 | 1 | 1 | +---+---+---+--------+ | 0 | 1 | 0 | 1 | +---+---+---+--------+ | 0 | 1 | 1 | 1 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+