(¬p ∨ ¬q ∨ ¬r) ∧ (¬p ∨ q ∨ ¬r) ∧ (p ∨ ¬q ∨ ¬r) ∧ (p ∨ ¬q ∨ r) ∧ (p ∨ q ∨ ¬r)
I could not simplify the above CNF. Could anyone please help me to simplify the statement
(¬p ∨ ¬q ∨ ¬r) ∧ (¬p ∨ q ∨ ¬r) ∧ (p ∨ ¬q ∨ ¬r) ∧ (p ∨ ¬q ∨ r) ∧ (p ∨ q ∨ ¬r)
I could not simplify the above CNF. Could anyone please help me to simplify the statement
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We can make a truth table for this:
If this truth table is correct, then an equivalent expression is:
We can factor out to simplify a bit...
We can also recognize that ¬q ∨ (p ∧ q) = ¬q ∨ p ...
This seems to match the truth table we got, so if that truth table is right, this expression is almost surely minimal or close to it. It only uses each variable once and so the only way that springs to mind to try to improve it is De Morgan, but it doesn't look like it reduces the number of operations even, so this might be as good as it gets.