Theorem sbcfng

Description: Distribute proper substitution through the function predicate with a domain. (Contributed by Alexander van der Vekens, 15-Jul-2018)

	Ref	Expression
Assertion	sbcfng	$$⊢X\in V\to \left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}FFnA\leftrightarrow ⦋X/x⦌FFn⦋X/x⦌A\right)$$

Proof

Step	Hyp	Ref	Expression
1		df-fn	$$⊢FFnA\leftrightarrow \left(\mathrm{Fun}F\wedge \mathrm{dom}F=A\right)$$
2	1	a1i	$$⊢X\in V\to \left(FFnA\leftrightarrow \left(\mathrm{Fun}F\wedge \mathrm{dom}F=A\right)\right)$$
3	2	sbcbidv	$$⊢X\in V\to \left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}FFnA\leftrightarrow \underset{\dot{}}{[}X/x\underset{\dot{}}{]}\left(\mathrm{Fun}F\wedge \mathrm{dom}F=A\right)\right)$$
4		sbcfung	$$⊢X\in V\to \left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}\mathrm{Fun}F\leftrightarrow \mathrm{Fun}⦋X/x⦌F\right)$$
5		sbceqg	$$⊢X\in V\to \left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}\mathrm{dom}F=A\leftrightarrow ⦋X/x⦌\mathrm{dom}F=⦋X/x⦌A\right)$$
6		csbdm	$$⊢⦋X/x⦌\mathrm{dom}F=\mathrm{dom}⦋X/x⦌F$$
7	6	eqeq1i	$$⊢⦋X/x⦌\mathrm{dom}F=⦋X/x⦌A\leftrightarrow \mathrm{dom}⦋X/x⦌F=⦋X/x⦌A$$
8	5 7	bitrdi	$$⊢X\in V\to \left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}\mathrm{dom}F=A\leftrightarrow \mathrm{dom}⦋X/x⦌F=⦋X/x⦌A\right)$$
9	4 8	anbi12d	$$⊢X\in V\to \left(\left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}\mathrm{Fun}F\wedge \underset{\dot{}}{[}X/x\underset{\dot{}}{]}\mathrm{dom}F=A\right)\leftrightarrow \left(\mathrm{Fun}⦋X/x⦌F\wedge \mathrm{dom}⦋X/x⦌F=⦋X/x⦌A\right)\right)$$
10		sbcan	$$⊢\underset{\dot{}}{[}X/x\underset{\dot{}}{]}\left(\mathrm{Fun}F\wedge \mathrm{dom}F=A\right)\leftrightarrow \left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}\mathrm{Fun}F\wedge \underset{\dot{}}{[}X/x\underset{\dot{}}{]}\mathrm{dom}F=A\right)$$
11		df-fn	$$⊢⦋X/x⦌FFn⦋X/x⦌A\leftrightarrow \left(\mathrm{Fun}⦋X/x⦌F\wedge \mathrm{dom}⦋X/x⦌F=⦋X/x⦌A\right)$$
12	9 10 11	3bitr4g	$$⊢X\in V\to \left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}\left(\mathrm{Fun}F\wedge \mathrm{dom}F=A\right)\leftrightarrow ⦋X/x⦌FFn⦋X/x⦌A\right)$$
13	3 12	bitrd	$$⊢X\in V\to \left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}FFnA\leftrightarrow ⦋X/x⦌FFn⦋X/x⦌A\right)$$