Mister Exam

# Expression (P⊕Q)∧((P→R)∨(Q→S))

### The solution

You have entered [src]
(p⊕q)∧((p⇒r)∨(q⇒s))
$$\left(\left(p \Rightarrow r\right) \vee \left(q \Rightarrow s\right)\right) \wedge \left(p ⊕ q\right)$$
Detail solution
$$p ⊕ q = \left(p \wedge \neg q\right) \vee \left(q \wedge \neg p\right)$$
$$p \Rightarrow r = r \vee \neg p$$
$$q \Rightarrow s = s \vee \neg q$$
$$\left(p \Rightarrow r\right) \vee \left(q \Rightarrow s\right) = r \vee s \vee \neg p \vee \neg q$$
$$\left(\left(p \Rightarrow r\right) \vee \left(q \Rightarrow s\right)\right) \wedge \left(p ⊕ q\right) = \left(p \wedge \neg q\right) \vee \left(q \wedge \neg p\right)$$
Simplification [src]
$$\left(p \wedge \neg q\right) \vee \left(q \wedge \neg p\right)$$
(p∧(¬q))∨(q∧(¬p))
Truth table
+---+---+---+---+--------+
| p | q | r | s | result |
+===+===+===+===+========+
| 0 | 0 | 0 | 0 | 0      |
+---+---+---+---+--------+
| 0 | 0 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 0 | 0 | 1 | 0 | 0      |
+---+---+---+---+--------+
| 0 | 0 | 1 | 1 | 0      |
+---+---+---+---+--------+
| 0 | 1 | 0 | 0 | 1      |
+---+---+---+---+--------+
| 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 | 0      |
+---+---+---+---+--------+
| 1 | 1 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 1 | 1 | 1 | 0 | 0      |
+---+---+---+---+--------+
| 1 | 1 | 1 | 1 | 0      |
+---+---+---+---+--------+
PDNF [src]
$$\left(p \wedge \neg q\right) \vee \left(q \wedge \neg p\right)$$
(p∧(¬q))∨(q∧(¬p))
PCNF [src]
$$\left(p \vee q\right) \wedge \left(\neg p \vee \neg q\right)$$
(p∨q)∧((¬p)∨(¬q))
CNF [src]
$$\left(p \vee q\right) \wedge \left(p \vee \neg p\right) \wedge \left(q \vee \neg q\right) \wedge \left(\neg p \vee \neg q\right)$$
(p∨q)∧(p∨(¬p))∧(q∨(¬q))∧((¬p)∨(¬q))
DNF [src]
$$\left(p \wedge \neg q\right) \vee \left(q \wedge \neg p\right)$$
(p∧(¬q))∨(q∧(¬p))