Mister Exam

Expression (p⇒(q⇒r))⇒((p⇒q)⇒(q⇒r))

    The solution

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