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]
    Already transformed to DNF
    $$\left(p \wedge \neg q\right) \vee \left(q \wedge \neg p\right)$$
    (p∧(¬q))∨(q∧(¬p))
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