If p → q is true, then ~ q → ~ p is __________ false. always sometimes never
always true
thnx
the answer is "never" false
\[\begin{array}{cccc}(p&\Rightarrow&q)&\implies (\neg p&\Rightarrow\neg q)\\\hline\top&&\top\\\top&&\bot\\\bot&&\top\\\bot&&\top\end{array}\]
fill in the rest of the truth table
\[\begin{array}{cccc}(p&\Rightarrow&q)&\implies& (\neg p&\Rightarrow&\neg q)\\\hline\top&\top&\top&&\bot&&\bot\\\top&\bot&\bot&&\bot&&\top\\\bot&\top&\top&&\top&&\bot\\\bot&\top&\bot&&\top&&\top\end{array}\]
\[\begin{array}{cccc}\color{red}(p&\color{blue}\Rightarrow&q\color{red})&\color{red}\implies& \color{red}(\neg p&\color{blue}\Rightarrow&\neg q\color{red})\\\hline\top&\color{blue}\top&\top&\color{red}\top&\bot&\color{blue}\top&\bot\\\top&\color{blue}\bot&\bot&\color{red}\top&\bot&\color{blue}\top&\top\\\bot&\color{blue}\top&\top&\color{red}\bot&\top&\color{blue}\bot&\bot\\\bot&\color{blue}\top&\bot&\color{red}\top&\top&\color{blue}\top&\top\end{array}\]
i think i read the question wrong
@UnkleRhaukus the question is ~q -> ~p. it is you that is wrong
sorry
the contrapositive has the same truth value as the implication
\[\begin{array}{cccc}\color{red}(p&\color{blue}\Rightarrow&q\color{red})&\color{red}\implies&\color{red}(\neg q&\color{blue}\Rightarrow&\neg p\color{red}) \\\hline\top&\color{blue}\top&\top&\color{red}\top&\bot&\color{blue}\top&\bot \\\top&\color{blue}\bot&\bot&\color{red}\top&\top&\color{blue}\bot&\bot \\\bot&\color{blue}\top&\top&\color{red}\top&\bot&\color{blue}\top&\top \\\bot&\color{blue}\top&\bot&\color{red}\top&\top&\color{blue}\top&\top\end{array}\]
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