Certainly! Let's construct the truth table for the logical statement [tex]\(\sim r \in (-q \vee p)\)[/tex].
First, let's define the logical operations and their symbols:
- [tex]\(\sim a\)[/tex] denotes the negation of [tex]\(a\)[/tex].
- [tex]\(|\sim a \vee p|\)[/tex] denotes the disjunction (OR) of [tex]\(\sim a\)[/tex] and [tex]\(p\)[/tex].
- [tex]\(\rightarrow\)[/tex] denotes implication.
In this case, we will enrich the truth table step-by-step and fill in the missing values.
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
D & [tex]$q$[/tex] & [tex]$r$[/tex] & [tex]$\sim r$[/tex] & [tex]$\sim a$[/tex] & [tex]$|\sim a \vee p|$[/tex] & [tex]$\sim r \rightarrow (-q \vee p)$[/tex] \\
\hline
[tex]$T$[/tex] & T & T & F & F & T & F \\
\hline
[tex]$T$[/tex] & T & F & T & T & T & T \\
\hline
\end{tabular}
1. For the first row:
- [tex]\(q = T\)[/tex]
- [tex]\(r = T\)[/tex]
- [tex]\(\sim r\)[/tex] is the negation of [tex]\(r\)[/tex], so [tex]\(\sim r = F\)[/tex]
- [tex]\(\sim a\)[/tex] is the negation of [tex]\(a\)[/tex], but it seems like 'a' is just used symbol here and doesn't affect the table since [tex]\(sim a = F\)[/tex] if 'a' is false.
- [tex]\(|\sim a \vee p|\)[/tex]. Since we do not know [tex]\(p\)[/tex]’s value, we assume it's T in implication generally for this Or case since we only care about q and r mostly. Hence, [tex]\(|\sim a \vee p| = T \)[/tex]
- [tex]\(\sim r \rightarrow (-q \vee p)\)[/tex]. Since [tex]\(- q \vee p\)[/tex] should cover False and False results in general cases and typically assuming [tex]\(sim r\)[/tex]-> requires implications check. Now since [tex]\(\sim r = F\)[/tex], a False premises makes whole implication true.
Hence [tex]\(\sim r \rightarrow (-q \vee p)= F\)[/tex]
Therefore:
Row 1: [tex]\(V = T\)[/tex] and [tex]\(\sim r \rightarrow (-q \vee p) = T\)[/tex]
Next step check for [tex]\(F\)[/tex] conditions similarizing concepts
2. For the second row:
- [tex]\(q = T\)[/tex]
- [tex]\(r = F\)[/tex]
- [tex]\(\sim r\)[/tex] is the negation of [tex]\(r\)[/tex], so [tex]\(\sim r = T\)[/tex]
- [tex]\(\sim a = T\)[/tex] as earlier already
- [tex]\(|\sim a \vee p|\)[/tex]: so we get assuming contradictionsn negations may satisfy only both Falsean dTrue premises carry. it as T mainly so
- [tex]\(\sim r \rightarrow (-q \vee p)\)[/tex]. Also test generally for complexity simplfiy as generally valid depending cases carry.
As usually for negated False premises = T makes
Therefore, for second row:
Generally [tex]\(V = T\)[/tex] and [tex]\( \sim D satisfies so \( = T\)[/tex]
\(\sim r \rightarrow\ (-q \vee p) True assertion
Thus verified.
