Sr Examen
Expresión (P→Q)→(R→S)
El profesor se sorprenderá mucho al ver tu solución correcta😉
⌨
Solución
Ha introducido
[src]
(p⇒q)⇒(r⇒s)
$$\left(p \Rightarrow q\right) \Rightarrow \left(r \Rightarrow s\right)$$
Solución detallada
$$p \Rightarrow q = q \vee \neg p$$
$$r \Rightarrow s = s \vee \neg r$$
$$\left(p \Rightarrow q\right) \Rightarrow \left(r \Rightarrow s\right) = s \vee \left(p \wedge \neg q\right) \vee \neg r$$
$$r \Rightarrow s = s \vee \neg r$$
$$\left(p \Rightarrow q\right) \Rightarrow \left(r \Rightarrow s\right) = s \vee \left(p \wedge \neg q\right) \vee \neg r$$
Tabla de verdad
+---+---+---+---+--------+ | p | q | r | s | result | +===+===+===+===+========+ | 0 | 0 | 0 | 0 | 1 | +---+---+---+---+--------+ | 0 | 0 | 0 | 1 | 1 | +---+---+---+---+--------+ | 0 | 0 | 1 | 0 | 0 | +---+---+---+---+--------+ | 0 | 0 | 1 | 1 | 1 | +---+---+---+---+--------+ | 0 | 1 | 0 | 0 | 1 | +---+---+---+---+--------+ | 0 | 1 | 0 | 1 | 1 | +---+---+---+---+--------+ | 0 | 1 | 1 | 0 | 0 | +---+---+---+---+--------+ | 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 | 0 | +---+---+---+---+--------+ | 1 | 1 | 1 | 1 | 1 | +---+---+---+---+--------+
FNC
[src]
$$\left(p \vee s \vee \neg r\right) \wedge \left(s \vee \neg q \vee \neg r\right)$$
(p∨s∨(¬r))∧(s∨(¬q)∨(¬r))
FNCD
[src]
$$\left(p \vee s \vee \neg r\right) \wedge \left(s \vee \neg q \vee \neg r\right)$$
(p∨s∨(¬r))∧(s∨(¬q)∨(¬r))
FND
[src]
Ya está reducido a FND
$$s \vee \left(p \wedge \neg q\right) \vee \neg r$$
s∨(¬r)∨(p∧(¬q))