Proofs:Biconditionals

A biconditional of two propositions P and Q takes the form " Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P } if and only if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q } ". This can also be written as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P \iff Q } , which is equivalent to $P\implies Q\land Q\implies P$. When proving a biconditional statement, we need to prove that $P\implies Q$ and $Q\implies P$ are true. Remember that the contrapositive of a conditional is logically equivalent to the conditional. Thus, " $P\implies Q$ and $Q\implies P$ " is logically equivalent to " $P\implies Q$ and $\neg P\implies \neg Q$ ", " $Q\implies P$ and $\neg Q\implies \neg P$ ", or " $\neg P\implies \neg Q$ and $\neg Q\implies \neg P$ ". Thus, we do have some options as to how to prove the two directions of a biconditional statement.

Example of a biconditional proof: "For $x\in \mathbb {Z} ,3x+3$ is odd if and only if $x$ is even". Let $P$ = "3x + 3 is odd" and $Q$ = "x is even".

If x is even, then 3x + 3 = 3(2k) + 3 for some integer k. 3(2k) + 3 = 3(2k + 1). 3 is odd and 2k+1 is odd, so their product is also odd. Thus, if x is even, 3x + 3 is odd. Therefore, the statement Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q \implies P } is true.

If x is odd, then 3x + 3 = 3(2k + 1) + 3 for some integer k. 3(2k + 1) + 3 = 3(2k + 1 + 1) = 3(2k + 2) = 3(2)(k+1). Thus 3x + 3 is a multiple of 2 and is even if x is odd. So, the statement Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \neg Q \implies \neg P } , which is logically equivalent to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P \implies Q } , is true.

Therefore, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P \iff Q } .

Resources

• Biconditional Statements, Mathematics LibreTexts
• Logical Biconditional, Wikipedia