We are given a truth table with columns for the propositions P, Q, and R. We need to complete the table by evaluating the expressions $P \rightarrow Q$, $Q \rightarrow R$, $P \land Q$, and $(P \rightarrow Q) \rightarrow (Q \rightarrow R)$ for all possible truth values of P, Q, and R.

Discrete MathematicsLogicTruth TablesPropositional Logic
2025/6/8

1. Problem Description

We are given a truth table with columns for the propositions P, Q, and R. We need to complete the table by evaluating the expressions PQP \rightarrow Q, QRQ \rightarrow R, PQP \land Q, and (PQ)(QR)(P \rightarrow Q) \rightarrow (Q \rightarrow R) for all possible truth values of P, Q, and R.

2. Solution Steps

We will use the following truth tables for implication (\rightarrow) and conjunction (\land):
PQP \rightarrow Q is false only when P is true and Q is false. Otherwise, it is true.
PQP \land Q is true only when both P and Q are true. Otherwise, it is false.
| P | Q | PQP \rightarrow Q | P | Q | PQP \land Q |
|---|---|--------------------|---|---|---------------|
| T | T | T | T | T | T |
| T | F | F | T | F | F |
| F | T | T | F | T | F |
| F | F | T | F | F | F |
Now, we can complete the truth table step by step.
Row 1: P=T, Q=T, R=T
PQ=TT=TP \rightarrow Q = T \rightarrow T = T
QR=TT=TQ \rightarrow R = T \rightarrow T = T
PQ=TT=TP \land Q = T \land T = T
(PQ)(QR)=TT=T(P \rightarrow Q) \rightarrow (Q \rightarrow R) = T \rightarrow T = T
Row 2: P=T, Q=T, R=F
PQ=TT=TP \rightarrow Q = T \rightarrow T = T
QR=TF=FQ \rightarrow R = T \rightarrow F = F
PQ=TT=TP \land Q = T \land T = T
(PQ)(QR)=TF=F(P \rightarrow Q) \rightarrow (Q \rightarrow R) = T \rightarrow F = F
Row 3: P=T, Q=F, R=T
PQ=TF=FP \rightarrow Q = T \rightarrow F = F
QR=FT=TQ \rightarrow R = F \rightarrow T = T
PQ=TF=FP \land Q = T \land F = F
(PQ)(QR)=FT=T(P \rightarrow Q) \rightarrow (Q \rightarrow R) = F \rightarrow T = T
Row 4: P=T, Q=F, R=F
PQ=TF=FP \rightarrow Q = T \rightarrow F = F
QR=FF=TQ \rightarrow R = F \rightarrow F = T
PQ=TF=FP \land Q = T \land F = F
(PQ)(QR)=FT=T(P \rightarrow Q) \rightarrow (Q \rightarrow R) = F \rightarrow T = T
Row 5: P=F, Q=T, R=T
PQ=FT=TP \rightarrow Q = F \rightarrow T = T
QR=TT=TQ \rightarrow R = T \rightarrow T = T
PQ=FT=FP \land Q = F \land T = F
(PQ)(QR)=TT=T(P \rightarrow Q) \rightarrow (Q \rightarrow R) = T \rightarrow T = T
Row 6: P=F, Q=T, R=F
PQ=FT=TP \rightarrow Q = F \rightarrow T = T
QR=TF=FQ \rightarrow R = T \rightarrow F = F
PQ=FT=FP \land Q = F \land T = F
(PQ)(QR)=TF=F(P \rightarrow Q) \rightarrow (Q \rightarrow R) = T \rightarrow F = F
Row 7: P=F, Q=F, R=T
PQ=FF=TP \rightarrow Q = F \rightarrow F = T
QR=FT=TQ \rightarrow R = F \rightarrow T = T
PQ=FF=FP \land Q = F \land F = F
(PQ)(QR)=TT=T(P \rightarrow Q) \rightarrow (Q \rightarrow R) = T \rightarrow T = T
Row 8: P=F, Q=F, R=F
PQ=FF=TP \rightarrow Q = F \rightarrow F = T
QR=FF=TQ \rightarrow R = F \rightarrow F = T
PQ=FF=FP \land Q = F \land F = F
(PQ)(QR)=TT=T(P \rightarrow Q) \rightarrow (Q \rightarrow R) = T \rightarrow T = T

3. Final Answer

Here is the completed truth table:
| P | Q | R | PQP \rightarrow Q | QRQ \rightarrow R | PQP \land Q | (PQ)(QR)(P \rightarrow Q) \rightarrow (Q \rightarrow R) |
|---|---|---|--------------------|--------------------|---------------|---------------------------------------------------|
| T | T | T | T | T | T | T |
| T | T | F | T | F | T | F |
| T | F | T | F | T | F | T |
| T | F | F | F | T | F | T |
| F | T | T | T | T | F | T |
| F | T | F | T | F | F | F |
| F | F | T | T | T | F | T |
| F | F | F | T | T | F | T |

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