Biconditional is Transitive/Formulation 1/Proof 1
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Theorem
- $p \iff q, q \iff r \vdash p \iff r$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \iff q$ | Premise | (None) | ||
2 | 2 | $q \iff r$ | Premise | (None) | ||
3 | 1 | $p \implies q$ | Biconditional Elimination: $\iff \EE_1$ | 1 | ||
4 | 2 | $q \implies r$ | Biconditional Elimination: $\iff \EE_1$ | 2 | ||
5 | 1, 2 | $p \implies r$ | Sequent Introduction | 1, 2 | Hypothetical Syllogism: Formulation 1 | |
6 | 1 | $q \implies p$ | Biconditional Elimination: $\iff \EE_2$ | 1 | ||
7 | 2 | $r \implies q$ | Biconditional Elimination: $\iff \EE_2$ | 2 | ||
8 | 1, 2 | $r \implies p$ | Sequent Introduction | 7, 6 | Hypothetical Syllogism: Formulation 1 | |
9 | 1, 2 | $p \iff r$ | Biconditional Introduction: $\iff \II$ | 5, 8 |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $4$ The Biconditional: Theorem $26$