Answer
See explanation
Work Step by Step
Below are **sample derivations** showing that each given expression is indeed a Boolean expression according to the stated grammar rules:
---
## Grammar Rules Recap
1. **Base Rule**:
Any single letter (from the given alphabet \(\{a,b,c,\dots,x,y,z\}\)) is a Boolean expression.
2. **Recursive Rules**:
- If \(P\) and \(Q\) are Boolean expressions, then \((P \land Q)\) is a Boolean expression.
- If \(P\) and \(Q\) are Boolean expressions, then \((P \lor Q)\) is a Boolean expression.
- If \(P\) is a Boolean expression, then \(\sim P\) (i.e.\ \(\lnot P\)) is a Boolean expression.
3. **Restriction**:
No other forms are allowed except those generated by the above rules.
---
## (a) \(\bigl(\sim p \;\lor\; (q \land (r \lor s))\bigr)\)
We want to show \(\bigl(\sim p \;\lor\; (q \land (r \lor s))\bigr)\) is a valid Boolean expression. Here is a step‐by‐step derivation:
1. By the **base rule**, \(p\), \(q\), \(r\), and \(s\) are each Boolean expressions (they are single letters in the alphabet).
2. From \(p\) being a Boolean expression, apply rule **II(c)** (\(\sim\) on an expression):
\[
p \quad\Longrightarrow\quad \sim p.
\]
So \(\sim p\) is a Boolean expression.
3. From \(r\) and \(s\) both being Boolean expressions, apply rule **II(b)** (\(\lor\) on two expressions):
\[
r, s \quad\Longrightarrow\quad (r \lor s).
\]
So \((r \lor s)\) is a Boolean expression.
4. Now \(q\) is a Boolean expression, and \((r \lor s)\) is a Boolean expression, so apply rule **II(a)** (\(\land\) on two expressions):
\[
q,\;(r \lor s) \quad\Longrightarrow\quad \bigl(q \land (r \lor s)\bigr).
\]
So \(\bigl(q \land (r \lor s)\bigr)\) is a Boolean expression.
5. Finally, we have \(\sim p\) and \(\bigl(q \land (r \lor s)\bigr)\) both as Boolean expressions, so apply rule **II(b)** (\(\lor\) on two expressions):
\[
\sim p,\;\bigl(q \land (r \lor s)\bigr)
\quad\Longrightarrow\quad
\bigl(\sim p \;\lor\; (q \land (r \lor s))\bigr).
\]
This shows \(\bigl(\sim p \;\lor\; (q \land (r \lor s))\bigr)\) is indeed a valid Boolean expression.
---
## (b) \(\bigl((p \lor q) \;\lor\; ((p \land \sim s) \land r)\bigr)\)
We want to show \(\bigl((p \lor q) \;\lor\; ((p \land \sim s) \land r)\bigr)\) is a valid Boolean expression. Derivation:
1. By the **base rule**, \(p\), \(q\), \(s\), and \(r\) are Boolean expressions.
2. From \(p\) and \(q\) both being Boolean expressions, apply rule **II(b)** (\(\lor\)):
\[
p,\;q \quad\Longrightarrow\quad (p \lor q).
\]
So \((p \lor q)\) is a Boolean expression.
3. From \(s\) being a Boolean expression, apply rule **II(c)** (\(\sim\)):
\[
s \quad\Longrightarrow\quad \sim s.
\]
So \(\sim s\) is a Boolean expression.
4. Now \(p\) and \(\sim s\) are Boolean expressions, so apply rule **II(a)** (\(\land\)):
\[
p,\;\sim s \quad\Longrightarrow\quad (p \land \sim s).
\]
So \((p \land \sim s)\) is a Boolean expression.
5. From \((p \land \sim s)\) and \(r\) being Boolean expressions, apply rule **II(a)** (\(\land\)) again:
\[
(p \land \sim s),\;r
\quad\Longrightarrow\quad
\bigl((p \land \sim s) \land r\bigr).
\]
So \(\bigl((p \land \sim s) \land r\bigr)\) is a Boolean expression.
6. Finally, we have \((p \lor q)\) and \(\bigl((p \land \sim s) \land r\bigr)\) as Boolean expressions, so apply rule **II(b)** (\(\lor\)):
\[
(p \lor q),\;\bigl((p \land \sim s) \land r\bigr)
\quad\Longrightarrow\quad
\bigl((p \lor q) \;\lor\; ((p \land \sim s) \land r)\bigr).
\]
This shows \(\bigl((p \lor q) \;\lor\; ((p \land \sim s) \land r)\bigr)\) is also a valid Boolean expression.
---
## Conclusion
By applying the base rule (single letters are expressions) and the recursive rules (building larger expressions from smaller ones via \(\land\), \(\lor\), and \(\sim\)), we have **derivations** for both
1. \(\bigl(\sim p \;\lor\; (q \land (r \lor s))\bigr)\), and
2. \(\bigl((p \lor q) \;\lor\; ((p \land \sim s) \land r)\bigr)\)
showing that each is indeed a **Boolean expression** in the sense of the given grammar.
