Self-Distributive Law for Conditional/Formulation 2
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Theorem
The following is known as the Self-Distributive Law:
- $\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }$
Proof
Proof of Forward Implication
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \implies \paren {q \implies r}$ | Assumption | (None) | ||
2 | 1 | $\paren {p \implies q} \implies \paren {p \implies r}$ | Sequent Introduction | 1 | Self-Distributive Law for Conditional: Formulation 1 | |
3 | $\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged |
$\blacksquare$
Proof of Reverse Implication
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\paren {p \implies q} \implies \paren {p \implies r}$ | Assumption | (None) | ||
2 | 1 | $p \implies \paren {q \implies r}$ | Sequent Introduction | 1 | Self-Distributive Law for Conditional: Formulation 1 | |
3 | $\paren {\paren {p \implies q} \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged |
$\blacksquare$
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | $\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | Theorem Introduction | (None) | Self-Distributive Law for Conditional: Forward Implication: Formulation 2 | ||
2 | $\paren {\paren {p \implies q} \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }$ | Theorem Introduction | (None) | Self-Distributive Law for Conditional: Reverse Implication: Formulation 2 | ||
3 | $\paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | Biconditional Introduction: $\iff \II$ | 1, 2 |
$\blacksquare$
Also see
- Conditional is not Right Self-Distributive where it is shown that while:
- $\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$
it is not the case that:
- $\paren {p \implies r} \implies \paren {q \implies r} \vdash \paren {p \implies q} \implies r$
Sources
- 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T106}$
- 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability ... (previous) ... (next): $\S 1.14$: Exercise $12 \ (8)$