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