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:

$\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} } $
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:

$\vdash \paren {\paren {p \implies q} \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} } $
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:

$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {\paren {p \implies q} \implies \paren {p \implies r} } $
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

$\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$

it is not the case that:

$\paren {p \implies r} \implies \paren {q \implies r} \vdash \paren {p \implies q} \implies r$


Sources