s is a set of strings recursively defined as ...
s is a set of strings recursively defined as follows. base case: every variable from the set {a, b, c, d, e, f} is in s. recursive rules: if x ∈ s and y ∈ s, then: 1. (x + y) ∈ s 2. x · y ∈ s indicate which expressions are in s. □ (a + b) · c · d □ (a + b + c) □ a · b □ a · (c + d)
Answer
# Explanation:
## Step1: Analyze base - case
The base - case gives that \(a,b,c,d,e,f\in S\).
## Step2: Analyze \((a + b)\cdot c\cdot d\)
Since \(a,b\in S\) (base - case), by recursive rule 1, \((a + b)\in S\). Since \(c\in S\) (base - case), then by recursive rule 2, \((a + b)\cdot c\in S\). Since \(d\in S\) (base - case), by recursive rule 2 again, \((a + b)\cdot c\cdot d\in S\).
## Step3: Analyze \((a + b + c)\)
Since \(a,b\in S\) (base - case), by recursive rule 1, \((a + b)\in S\). Since \(c\in S\) (base - case), by recursive rule 1 again, \((a + b)+c=(a + b + c)\in S\).
## Step4: Analyze \(a\cdot b\)
Since \(a,b\in S\) (base - case), by recursive rule 2, \(a\cdot b\in S\).
## Step5: Analyze \(a\cdot(c + d)\)
Since \(c,d\in S\) (base - case), by recursive rule 1, \((c + d)\in S\). Since \(a\in S\) (base - case), by recursive rule 2, \(a\cdot(c + d)\in S\).
# Answer:
\((a + b)\cdot c\cdot d\), \((a + b + c)\), \(a\cdot b\), \(a\cdot(c + d)\) are all in \(S\).