Principle of Commutation/Formulation 2

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Theorem

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


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:

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

$\blacksquare$


By the tableau method of natural deduction:

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

$\blacksquare$


Sources