Using Boolean algebra simplify the statement ¬(𝑟 → 𝑠) → (¬𝑟)
Using Boolean algebra let us simplify the statement:
¬(𝑟→𝑠)→(¬𝑟)=¬(¬(𝑟→𝑠))∨(¬𝑟)=(𝑟→𝑠)∨(¬𝑟)=(¬𝑟∨𝑠)∨(¬𝑟)=(s∨¬𝑟)∨(¬𝑟)=s∨(¬𝑟∨¬𝑟)=s∨¬𝑟=¬𝑟∨s=𝑟→𝑠.¬(𝑟 → 𝑠) → (¬𝑟) =¬(¬(𝑟 → 𝑠))\lor (¬𝑟) =(𝑟 → 𝑠)\lor (¬𝑟) =(¬𝑟\lor 𝑠)\lor (¬𝑟) = (s\lor ¬𝑟)\lor (¬𝑟) =s\lor( ¬𝑟\lor ¬𝑟) =s\lor ¬𝑟=¬𝑟\lor s=𝑟 → 𝑠.¬(r→s)→(¬r)=¬(¬(r→s))∨(¬r)=(r→s)∨(¬r)=(¬r∨s)∨(¬r)=(s∨¬r)∨(¬r)=s∨(¬r∨¬r)=s∨¬r=¬r∨s=r→s.
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