First, we need to understand how to solve the truth table
We have the variables p,r, and s
Where:
∧ is the symbol for disjunction = either or both
→ is the symbol for conditional = if .... so
Now, we can solve the truth table:
T = means true
F = means false
We need to find each different case:
prsp ∧ (r → s)
First, we solve (r → s)
If r is true and s is true, the result is true:
r s (r → s)
T T T
T F F
F T T
F F T
Now, for p ∧ (r → s)
p r s p ∧ (r → s)
First case: When
p r s
T T T
T ∧ (T → T)
where (T → T) =T
Then :
T ∧ (T)
The result is T
Therefore:
prsp ∧ (r → s)
TTT T
Second case:
When
p r s
T T F
where (r → s) = (T → F) = F
Then
p ∧ (r → s) = T ∧ F = F
Therefore:
prsp ∧ (r → s)
TTF F
Third case:
When
p r s
T F T
Where (r → s) = (F → T) = T
Then
p ∧ (r → s) = T ∧ T = T
Therefore:
prsp ∧ (r → s)
TFT T
Fourth case:
When
prs
TFF
Where (r → s) = (F → F) = T
Then
p ∧ (r → s) = T ∧ T = T
Therefore:
prsp ∧ (r → s)
TFF T
Fifth case:
Where
p r s
F T T
Where (r → s) = (T → T) = T
Then
p ∧ (r → s) = F ∧ T = F
Therefore:
prsp ∧ (r → s)
FTT F
Sixth case:
Where
p r s
F T F
Where (r → s) = (T → F) = F
Then
p ∧ (r → s) = F ∧ F = F
Therefore:
prsp ∧ (r → s)
FTF F
Seventh case:
Where
p r s
F F T
Where (r → s) = (F→ T ) = T
Then
p ∧ (r → s) = F ∧ T = F
Therefore:
prsp ∧ (r → s)
FFT F
Last case
Where
p r s
F F F
Where (r → s) = (F→F) = T
Then
p ∧ (r → s) = F ∧ T = F
Therefore:
prsp ∧ (r → s)
FFF F
Hence, we can get the whole truth table:
prsp ∧ (r → s)
TTT T
TTF F
TFT T
TFFT
FTTF
FTFF
FFTF
FFF F