CMSC250
Logic
Logic
Necessary and Sufficient
Recall: \(p \Rightarrow q\)
\(p \Rightarrow q\)
\(q\) is necessary for \(p\)
It is impossible to have \(p\) without \(q\)
\(p \Rightarrow q\)
It is impossible to have \(p\) without \(q\)
It is necessary for the sidewalk to be wet if it rained
\(p \Rightarrow q\)
It is impossible to have \(p\) without \(q\)
If the sidewalk is not wet, then it could not have rained
\(p \Rightarrow q\)
It is impossible to have \(p\) without \(q\)
If the sidewalk is not wet, then it could not have rained
If it rained, then the sidewalk is wet
\(p \Rightarrow q\)
\(p\) is sufficient for \(q\)
\(p \Rightarrow q\)
\(p\) is sufficient for \(q\)
Knowing it rained is sufficient enough to conclude the sidewalk is wet
\(p \Rightarrow q\)
\(p\) is sufficient for \(q\)
If it rained, then the sidewalk is wet
If \(p\) is true, then \(q\) is true
Recap: Logical Equivalence
~\((p \lor q) \equiv (\)~\(p)\land (\)~\(q)\)
~\((p \lor q) \equiv (\)~\(p)\land (\)~\(q)\)
We can show equivalence via Truth Table
| \(p\) | \(q\) | ~\(p\) | ~\(q\) | \(p \lor q\) | ~\((p \lor q)\) | \((\)~\(p)\land(\)~\(q)\) |
|---|---|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 | 0 | 0 |
| 1 | 1 | 0 | 0 | 1 | 0 | 0 |
~\((p \lor q) \equiv (\)~\(p)\land (\)~\(q)\)
~\((p \land q) \equiv (\)~\(p)\lor (\)~\(q)\)
Note: These together are known as DeMorgan's Laws
What other laws exist?
What other laws exist?
\(p \land ({\sim}p)\) (Contradiction)
\(p \lor ({\sim}p)\) (Tautology)
\(p \lor q \equiv q \lor p\)
\(p \lor (q \lor r) \equiv (p \lor q) \lor r\)
\(p \lor q \equiv q \lor p\)
| \(p\) | \(q\) | \(p \lor q\) | \(q \lor p\) |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 |
Commutative law
\(p \lor (q \lor r) \equiv (p \lor q) \lor r\)
| \(p\) | \(q\) | \(r\) | \(p \lor q\) | \(q \lor r\) | \(p \lor (q \lor r)\) | \((p \lor q) \lor r\) |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 1 |
| 1 | 0 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Associative law
Rules of Equivalence
Note: Variables can stand for longer statements
\((a \land b \lor c \land d) \lor 1 \equiv 1\)
Prove: \(({\sim}(({\sim} p) \land q)) \land (p \lor q) \equiv p\)
Prove: \(({\sim}(({\sim} p) \land q)) \land (p \lor q) \equiv p\)
| \(p\) | \(q\) | \({\sim}p\) | \(({\sim}p) \land q\) | \({\sim}(({\sim}p) \land q)\) | \(p\lor q\) | \(({\sim}(({\sim} p) \land q)) \land (p \lor q)\) |
|---|---|---|---|---|---|---|
| 0 | 0 | 1 | 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 | 0 | 1 | 0 |
| 1 | 0 | 0 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 1 | 1 |
Prove: \(({\sim}(({\sim} p) \land q)) \land (p \lor q) \equiv p\)
\[\begin{array}{rll} & \fragment{1}{({\sim}(({\sim} p) \land q)) \land (p \lor q)}&\\ \fragment{2}{\equiv} & \fragment{2}{({\sim}({\sim} p) \lor ({\sim}q)) \land (p \lor q)} & \fragment{3}{\text{De Morgan's}}\\ \fragment{4}{\equiv} & \fragment{4}{(p \lor ({\sim}q)) \land (p \lor q)} & \fragment{5}{\text{Double Negation}}\\ \fragment{6}{\equiv} & \fragment{6}{(p \lor (({\sim}q)) \land q)} & \fragment{7}{\text{Distributive}}\\ \fragment{8}{\equiv} & \fragment{8}{p \lor 0} & \fragment{9}{\text{Negation}}\\ \fragment{10}{\equiv} & \fragment{10}{p} & \fragment{11}{\text{Identity}}\\ \end{array} \]
Prove \({\sim}(({\sim} p) \lor ({\sim}(q \land p))) \equiv p \land q\)
Prove \({\sim}(({\sim} p) \lor ({\sim}(q \land p))) \equiv p \land q\)
\[\begin{array}{rll} & \fragment{1}{{\sim}(({\sim} p) \lor ({\sim}(q \land p)))}& \\ \fragment{2}{\equiv} &\fragment{2}{({\sim}({\sim} p)) \land ({\sim}({\sim}(q \land p)))} &\fragment{3}{\text{DeMorgan's Law}}\\ \fragment{4}{\equiv} &\fragment{4}{p \land (q \land p)} &\fragment{5}{\text{Double Negation Law}} \\ \fragment{6}{\equiv} &\fragment{6}{p \land (p \land q)} &\fragment{7}{\text{Communitive Law}} \\ \fragment{8}{\equiv} &\fragment{8}{(p \land p) \land q} &\fragment{9}{\text{Associative Law}} \\ \fragment{10}{\equiv} &\fragment{10}{p \land q} &\fragment{11}{ \text{Idempotent Law}} \\ \end{array}\]
Prove \({\sim} (p \lor (({\sim}p) \land q)) \equiv {\sim}p \land {\sim}q\)
Prove \({\sim} (p \lor (({\sim}p) \land q)) \equiv {\sim}p \land {\sim}q\)
\(\begin{array}{rll} & \fragment{1}{{\sim} (p \lor (({\sim}p) \land q))}& \\ \fragment{2}{\equiv} &\fragment{2}{({\sim}p) \land ({\sim}(({\sim}p) \land q))}& \fragment{3}{\text{DeMorgan's Law}}\\ \fragment{4}{\equiv} &\fragment{4}{({\sim}p) \land (({\sim}({\sim}p)) \lor ({\sim} q))}& \fragment{5}{\text{DeMorgan's Law}}\\ \fragment{6}{\equiv} &\fragment{6}{({\sim}p) \land (p \lor ({\sim} q))}& \fragment{7}{\text{Double Negation Law}}\\ \fragment{8}{\equiv} &\fragment{8}{(({\sim}p) \land p) \lor (({\sim} p) \land ({\sim} q))}& \fragment{9}{\text{Distributive Law}}\\ \fragment{10}{\equiv} &\fragment{10}{0 \lor ({\sim} p \land {\sim} q)}& \fragment{11}{\text{Negation Law}}\\ \fragment{12}{\equiv} &\fragment{12}{({\sim} p) \land ({\sim} q)}& \fragment{13}{\text{Identity Law}}\\ \end{array}\)
Arguments
An Argument consists of premises (hypotheses) and a conclusion
An Argument consists of premises (hypotheses) and a conclusion
\[ \begin{array}{c@{\,}l@{}} & q \\ & p \lor q \\\hline \therefore & p \rightarrow q \end{array} \]
An Argument consists of premises (hypotheses) and a conclusion
Arguments can be valid, both, or neither
An Argument is valid if when the premises are true, the conclusion must also be true
An Argument is valid if when the premises are true, the conclusion must also be true
\[ \begin{array}{c@{\,}l@{}} & q \\ & p \lor q \\\hline \therefore & p \rightarrow q \end{array} \]
If an Argument is valid, then \((h_1 \land h_2 \land \dots) \Rightarrow c\)
If an Argument is valid, then \((h_1 \land h_2 \land \dots) \Rightarrow c\)
\[ \begin{array}{c@{\,}l@{}} & q \\ & p \lor q \\\hline \therefore & p \rightarrow q \end{array} \]
| \(p\) | \(q\) | \(p \lor q\) | \(p \rightarrow q\) |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 1 |
If an Argument is valid, then \((h_1 \land h_2 \land \dots) \Rightarrow c\)
\[ \begin{array}{c@{\,}l@{}} & p \\ & p \lor q \\\hline \therefore & p \rightarrow q \end{array} \]
| \(p\) | \(q\) | \(p \lor q\) | \(p \rightarrow q\) |
|---|---|---|---|
| 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 1 |
An Argument is sound if the argument is valid and all premises are actually True
If an Argument is sound, then \((h_1\land h_2 \land \dots) \land ((h_1 \land h_2 \land \dots)\Rightarrow c)\)
An Argument is sound if the argument is valid and all premises are actually True
If an Argument is sound, then \((h_1\land h_2 \land \dots) \land ((h_1 \land h_2 \land \dots)\Rightarrow c)\)
\[ \begin{array}{c@{\,}l@{}} & \text{I am older than 18}\\ & \text{If I am older than 18, then I am not a teenager}\\\hline \therefore & \text{I am not a teenager} \end{array} \]
An Argument is sound if the argument is valid and all premises are actually True
If an Argument is sound, then \((h_1\land h_2 \land \dots) \land ((h_1 \land h_2 \land \dots)\Rightarrow c)\)
\[ \begin{array}{c@{\,}l@{}} & \text{I am 18}\\ & \text{If I am 18, then I am a teenager}\\\hline \therefore & \text{I am a teenager} \end{array} \]
\[ \begin{array}{c@{\,}l@{}} & \text{I am 18}\\ & \text{If I am 18, then I am a teenager}\\\hline \therefore & \text{I am a teenager} \end{array} \]
| I am 18 | I am teenager | If I am 18, then I am a teenager |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |