Proof by Cases/Formulation 1/Forward Implication
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Theorem
- $\paren {p \implies r} \land \paren {q \implies r} \vdash \paren {p \lor q} \implies r$
Proof 1
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\paren {p \implies r} \land \paren {q \implies r}$ | Premise | (None) | ||
2 | 1 | $p \implies r$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
3 | 1 | $q \implies r$ | Rule of Simplification: $\land \EE_2$ | 1 | ||
4 | 1 | $p \lor q \implies r \lor r$ | Sequent Introduction | 2, 3 | Constructive Dilemma | |
5 | 5 | $p \lor q$ | Assumption | (None) | ||
6 | 1, 5 | $r \lor r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 4, 5 | ||
7 | 1, 5 | $r$ | Sequent Introduction | 6 | Rule of Idempotence: Disjunction | |
8 | 1 | $p \lor q \implies r$ | Rule of Implication: $\implies \II$ | 5 – 7 | Assumption 5 has been discharged |
$\blacksquare$
Proof 2
From the Constructive Dilemma we have:
- $p \implies q, r \implies s \vdash p \lor r \implies q \lor s$
from which, changing the names of letters strategically:
- $p \implies r, q \implies r \vdash p \lor q \implies r \lor r$
From the Rule of Idempotence we have:
- $r \lor r \vdash r$
and the result follows by Hypothetical Syllogism.
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $3$ Conjunction and Disjunction: Exercise $1 \ \text{(f)}$
- 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability ... (previous) ... (next): $\S 1.13$: Tableau Problems (TAB1): CASES