# Principle of Commutation/Formulation 2

## Theorem

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

This can of course be expressed as two separate theorems:

### Forward Implication

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

### Reverse Implication

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

## Proof

### Proof of Forward Implication

By the tableau method of natural deduction:

$\vdash \left({p \implies \left({q \implies r}\right)}\right) \implies \left({q \implies \left({p \implies r}\right)}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies \left({q \implies r}\right)$ Assumption (None)
2 1 $q \implies \left({p \implies r}\right)$ Sequent Introduction 1 Principle of Commutation: Forward Implication: Formulation 1
3 $\left({p \implies \left({q \implies r}\right)}\right) \implies \left({q \implies \left({p \implies r}\right)}\right)$ Rule of Implication: $\implies \mathcal I$ 1 – 2 Assumption 1 has been discharged

$\blacksquare$

### Proof of Reverse Implication

By the tableau method of natural deduction:

$\vdash \paren {q \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }$
Line Pool Formula Rule Depends upon Notes
1 1 $q \implies \paren {p \implies r}$ Assumption (None)
2 1 $p \implies \paren {q \implies r}$ Sequent Introduction 1 Principle of Commutation: Reverse Implication: Formulation 1
3 $\paren {q \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }$ Rule of Implication: $\implies \mathcal I$ 1 – 2 Assumption 1 has been discharged

$\blacksquare$

By the tableau method of natural deduction:

$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} }$
Line Pool Formula Rule Depends upon Notes
1 $\paren {p \implies \paren {q \implies r} } \implies \paren {q \implies \paren {p \implies r} }$ Theorem Introduction (None) Principle of Commutation: Forward Implication: Formulation 2
2 $\paren {q \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }$ Theorem Introduction (None) Principle of Commutation: Reverse Implication: Formulation 2
3 $\paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} }$ Biconditional Introduction: $\iff \mathcal I$ 1, 2

$\blacksquare$