show that ~Q,P—>Q=>~P in mathematical foundations of computer science
Let us construct a truth table for the compound proposition (∼Q∧(P→Q))→∼P:(\sim Q\land (P\to Q))\to \sim P:(∼Q∧(P→Q))→∼P:
PQP→Q∼Q∼P∼Q∧(P→Q)(∼Q∧(P→Q))→∼P0011111011010110010011110001\begin{array}{||c|c||c|c|c|c|c||} \hline\hline P & Q & P\to Q & \sim Q & \sim P & \sim Q\land (P\to Q) & (\sim Q\land (P\to Q))\to \sim P \\ \hline\hline 0 & 0 & 1 & 1 & 1 & 1 & 1\\ \hline 0 & 1 & 1 & 0 & 1 & 0 & 1 \\ \hline 1 & 0 & 0 & 1 & 0 & 0 & 1\\ \hline 1 & 1 & 1 & 0 & 0 & 0 & 1\\ \hline\hline \end{array}P0011Q0101P→Q1101∼Q1010∼P1100∼Q∧(P→Q)1000(∼Q∧(P→Q))→∼P1111
It follows that the formula (∼Q∧(P→Q))→∼P(\sim Q\land (P\to Q))\to \sim P(∼Q∧(P→Q))→∼P is a tautology, and hence ∼Q,P→Q⊢ ∼P.\sim Q, P\to Q\vdash\ \sim P.∼Q,P→Q⊢ ∼P.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment