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Now, $\scriptstyle {Q_{1}|\pi {\mathsf {I}}\pi }$ fulfils (α) and (β). For (α) $\scriptstyle {\pi }$ being the complex expression $\scriptstyle {t{\mathsf {I}}t|t}$, is a case of the form $\scriptstyle {q|r}$, and (β) we have, by (c) above, $\scriptstyle {\pi |\pi /Q_{1}}$$\scriptstyle {Q_{1}/\pi |\pi ,}$ and by (a) $\scriptstyle {\pi {\mathsf {I}}\pi |Q_{1}}$.

To obtain the strictest development of the proof we have only to write $\scriptstyle {Q_{1}/\pi |\pi }$ for $\scriptstyle {P}$ and $\scriptstyle {\pi |\pi /Q_{1}}$ for $\scriptstyle {T}$ all through the preceding argument.

Permutation,

$\scriptstyle {s|p}$$\scriptstyle {p|s}$

$\scriptstyle {\left[{\text{Gives }}s\vee p.\supset .p\vee s{\text{ by }}{\frac {p|p\quad s|s}{p\qquad s}}.\right]}$

Dem.: Prop. $\scriptstyle {\frac {p\quad p\quad p}{p\quad q\quad r}}$, Id., and Rule.

Tautology,	$\scriptstyle {p/p|p/p{\mathsf {I}}p/p}$
	i.e. $\scriptstyle {p\vee p.\supset .p}$

Dem.: Id. $\scriptstyle {\frac {p/p}{p}}$, Perm., and Rule.

Addition,

$\scriptstyle {s}$$\scriptstyle {p|s/s}$

$\scriptstyle {\left[{\text{Gives }}s.\supset .p\vee s{\text{ by }}{\frac {p/p}{p}}.\right]}$

Dem.: By Perm. (twice), $\scriptstyle {p|s/s}$$\scriptstyle {s/s|p}$ (a)

By Prop. $\scriptstyle {{\frac {{\overline {p|s/s}}\quad s/s\quad p\quad s}{\;\;p\qquad \;\,q\quad \;\;r\quad s}},\;\vdash (a),\;\vdash }$ Id.^[1], $\scriptstyle {p|s/s}$$\scriptstyle {s}$

By Perm., result.

Return from Generalised Implication $\scriptstyle {P|\pi /Q}$ to $\scriptstyle {P|Q/Q}$.

Lemma,

$\scriptstyle {p/p}$$\scriptstyle {s/p}$

Dem.: By Perm. (twice), $\scriptstyle {s/p}$$\scriptstyle {p/s}$ (a)

By Prop. $\scriptstyle {{\frac {{\overline {s/p}}\quad p\quad s\quad u}{\;p\quad \;\,q\quad r\quad s}},\;\vdash }$(a),

$\scriptstyle {u|p}$$\scriptstyle {s/p}$$\scriptstyle {u}$

Write $\scriptstyle {p/p}$ for $\scriptstyle {u}$: by Id. and Perm. (twice), result.

1. $\scriptstyle {\vdash }$(a) means the use of the Rule to pass from $\scriptstyle {a}$ to $\scriptstyle {b}$ in $\scriptstyle {a|s/b}$.