# Rule of Exportation/Forward Implication/Formulation 1/Proof

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## Theorem

- $\left ({p \land q}\right) \implies r \vdash p \implies \left ({q \implies r}\right)$

## Proof

By the tableau method of natural deduction:

Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|

1 | 1 | $\left ({p \land q}\right) \implies r$ | Premise | (None) | ||

2 | 2 | $p$ | Assumption | (None) | ||

3 | 3 | $q$ | Assumption | (None) | ||

4 | 2, 3 | $p \land q$ | Rule of Conjunction: $\land \mathcal I$ | 2, 3 | ||

5 | 1, 2, 3 | $r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 4 | ||

6 | 1, 2 | $q \implies r$ | Rule of Implication: $\implies \mathcal I$ | 3 – 5 | Assumption 3 has been discharged | |

7 | 1 | $p \implies \left ({q \implies r}\right)$ | Rule of Implication: $\implies \mathcal I$ | 2 – 6 | Assumption 2 has been discharged |

$\blacksquare$

## Sources

- 1965: E.J. Lemmon:
*Beginning Logic*... (previous) ... (next): $\S 1.3$: Conjunction and Disjunction: Theorem $13$ - 2000: Michael R.A. Huth and Mark D. Ryan:
*Logic in Computer Science: Modelling and reasoning about systems*... (previous) ... (next): $\S 1.2.1$: Rules for natural deduction: Example $1.13$