Biconditional Math Example 4

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Example 4

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Verify that pโ‡”qโ‰ก(pโ‡’q)โˆง(qโ‡’p)p \Leftrightarrow q \equiv (p \Rightarrow q) \land (q \Rightarrow p) using a truth table.

Solution

  1. 1
    Compute pโ‡’qp \Rightarrow q: T,F,T,TT,F,T,T. Compute qโ‡’pq \Rightarrow p: T,T,F,TT,T,F,T.
  2. 2
    (pโ‡’q)โˆง(qโ‡’p)(p \Rightarrow q) \land (q \Rightarrow p): TโˆงT,FโˆงT,TโˆงF,TโˆงT=T,F,F,TT\land T, F\land T, T\land F, T\land T = T,F,F,T.
  3. 3
    pโ‡”qp \Leftrightarrow q column is also T,F,F,TT,F,F,T. The columns match.

Answer

pโ‡”qโ‰ก(pโ‡’q)โˆง(qโ‡’p)p \Leftrightarrow q \equiv (p \Rightarrow q) \land (q \Rightarrow p)
The biconditional is logically equivalent to requiring both directions of implication to hold simultaneously. This decomposition is how biconditionals are typically proved.

About Biconditional

A biconditional Pโ†”QP \leftrightarrow Q is true when PP and QQ have the same truth value โ€” both true or both false.

Learn more about Biconditional โ†’

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