Sr Examen
Expresión not(p⇒q)⇒notpv(q&r)
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
(¬(p⇒q))⇒((¬p)∨(q∧r))
$$p \not\Rightarrow q \Rightarrow \left(\left(q \wedge r\right) \vee \neg p\right)$$
Solución detallada
$$p \Rightarrow q = q \vee \neg p$$
$$p \not\Rightarrow q = p \wedge \neg q$$
$$p \not\Rightarrow q \Rightarrow \left(\left(q \wedge r\right) \vee \neg p\right) = q \vee \neg p$$
$$p \not\Rightarrow q = p \wedge \neg q$$
$$p \not\Rightarrow q \Rightarrow \left(\left(q \wedge r\right) \vee \neg p\right) = q \vee \neg p$$
Tabla de verdad
+---+---+---+--------+ | p | q | r | result | +===+===+===+========+ | 0 | 0 | 0 | 1 | +---+---+---+--------+ | 0 | 0 | 1 | 1 | +---+---+---+--------+ | 0 | 1 | 0 | 1 | +---+---+---+--------+ | 0 | 1 | 1 | 1 | +---+---+---+--------+ | 1 | 0 | 0 | 0 | +---+---+---+--------+ | 1 | 0 | 1 | 0 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+