Theorem sbcfg

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

	Ref	Expression
Assertion	sbcfg	$$⊢X\in V\to \left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}F:A⟶B\leftrightarrow ⦋X/x⦌F:⦋X/x⦌A⟶⦋X/x⦌B\right)$$

Proof

Step	Hyp	Ref	Expression
1		df-f	$$⊢F:A⟶B\leftrightarrow \left(FFnA\wedge \mathrm{ran}F\subseteq B\right)$$
2	1	a1i	$$⊢X\in V\to \left(F:A⟶B\leftrightarrow \left(FFnA\wedge \mathrm{ran}F\subseteq B\right)\right)$$
3	2	sbcbidv	$$⊢X\in V\to \left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}F:A⟶B\leftrightarrow \underset{\dot{}}{[}X/x\underset{\dot{}}{]}\left(FFnA\wedge \mathrm{ran}F\subseteq B\right)\right)$$
4		sbcfng	$$⊢X\in V\to \left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}FFnA\leftrightarrow ⦋X/x⦌FFn⦋X/x⦌A\right)$$
5		sbcssg	$$⊢X\in V\to \left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}\mathrm{ran}F\subseteq B\leftrightarrow ⦋X/x⦌\mathrm{ran}F\subseteq ⦋X/x⦌B\right)$$
6		csbrn	$$⊢⦋X/x⦌\mathrm{ran}F=\mathrm{ran}⦋X/x⦌F$$
7	6	sseq1i	$$⊢⦋X/x⦌\mathrm{ran}F\subseteq ⦋X/x⦌B\leftrightarrow \mathrm{ran}⦋X/x⦌F\subseteq ⦋X/x⦌B$$
8	5 7	bitrdi	$$⊢X\in V\to \left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}\mathrm{ran}F\subseteq B\leftrightarrow \mathrm{ran}⦋X/x⦌F\subseteq ⦋X/x⦌B\right)$$
9	4 8	anbi12d	$$⊢X\in V\to \left(\left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}FFnA\wedge \underset{\dot{}}{[}X/x\underset{\dot{}}{]}\mathrm{ran}F\subseteq B\right)\leftrightarrow \left(⦋X/x⦌FFn⦋X/x⦌A\wedge \mathrm{ran}⦋X/x⦌F\subseteq ⦋X/x⦌B\right)\right)$$
10		sbcan	$$⊢\underset{\dot{}}{[}X/x\underset{\dot{}}{]}\left(FFnA\wedge \mathrm{ran}F\subseteq B\right)\leftrightarrow \left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}FFnA\wedge \underset{\dot{}}{[}X/x\underset{\dot{}}{]}\mathrm{ran}F\subseteq B\right)$$
11		df-f	$$⊢⦋X/x⦌F:⦋X/x⦌A⟶⦋X/x⦌B\leftrightarrow \left(⦋X/x⦌FFn⦋X/x⦌A\wedge \mathrm{ran}⦋X/x⦌F\subseteq ⦋X/x⦌B\right)$$
12	9 10 11	3bitr4g	$$⊢X\in V\to \left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}\left(FFnA\wedge \mathrm{ran}F\subseteq B\right)\leftrightarrow ⦋X/x⦌F:⦋X/x⦌A⟶⦋X/x⦌B\right)$$
13	3 12	bitrd	$$⊢X\in V\to \left(\underset{\dot{}}{[}X/x\underset{\dot{}}{]}F:A⟶B\leftrightarrow ⦋X/x⦌F:⦋X/x⦌A⟶⦋X/x⦌B\right)$$