Calculating Truth Tables for Propositions

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SUMMARY

This discussion focuses on calculating truth tables for the propositions $$(p \land q) \lor (\lnot p \land q) \to q$$ and $$p \land \lnot q \to r$$. The truth tables provided demonstrate the logical relationships between the variables p, q, and r, confirming that the propositions hold true under specific conditions. The calculations show that when p and q are both true or both false, the implications are valid, while the truth of the second proposition depends on the value of r when p is true and q is false.

PREREQUISITES
  • Understanding of propositional logic
  • Familiarity with truth tables
  • Knowledge of logical operators such as conjunction (AND), disjunction (OR), and negation (NOT)
  • Basic skills in mathematical notation
NEXT STEPS
  • Study the construction of truth tables for more complex logical expressions
  • Learn about logical equivalences and how they apply to propositions
  • Explore implications in propositional logic and their interpretations
  • Investigate the use of software tools for automated truth table generation
USEFUL FOR

Students of mathematics, logic enthusiasts, and anyone interested in understanding the foundations of propositional logic and truth table calculations.

evinda
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Hello! (Wave)

I want to calculate the truth tables of the following propositions:

$$(p \land q) \lor (\lnot p \land q) \to q \\ p \land \lnot q \to r$$

I have done the following:\begin{equation*}
\begin{array}{c|c|c|c|c}
p & q & p \land q & \lnot p \land q & (p \land q) \lor (\lnot p \land q) \to q \\
\hline
1 & 1 & 1 & 0 & 1 \\
1 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 1 & 1
\end{array}
\end{equation*}

and

\begin{equation*}
\begin{array}{c|c|c|c}
p & q & \lnot q & p \land \lnot q \\
\hline
1 & 1 & 0 & 0 \\
1 & 0 & 1 & 1 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0
\end{array}
\end{equation*}

If $p=q=1$ and $r=0$, then $p \land \lnot q \to r$ is true, and the same holds if $r=1$. The same holds when $p=q=0$ and $p=0$, $q=1$.

If $p=1$ and $q=0$, then if $r=0$ then $p \land \lnot q \to r$ is false, and if $r=1$ then it is true.
Is everything right? Or have I done something wrong? :unsure:
 
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Hi evinda,

That looks correct
 
castor28 said:
Hi evinda,

That looks correct

Nice, thank you! :)
 

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