Theorem sbccomlem

Description: Lemma for sbccom . (Contributed by NM, 14-Nov-2005) (Revised by Mario Carneiro, 18-Oct-2016)

	Ref	Expression
Assertion	sbccomlem	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\underset{\dot{}}{[}B/y\underset{\dot{}}{]}\phi \leftrightarrow \underset{\dot{}}{[}B/y\underset{\dot{}}{]}\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi$$

Proof

Step	Hyp	Ref	Expression
1		excom	$$⊢\exists x\exists y\left(x=A\wedge \left(y=B\wedge \phi \right)\right)\leftrightarrow \exists y\exists x\left(x=A\wedge \left(y=B\wedge \phi \right)\right)$$
2		exdistr	$$⊢\exists x\exists y\left(x=A\wedge \left(y=B\wedge \phi \right)\right)\leftrightarrow \exists x\left(x=A\wedge \exists y\left(y=B\wedge \phi \right)\right)$$
3		an12	$$⊢\left(x=A\wedge \left(y=B\wedge \phi \right)\right)\leftrightarrow \left(y=B\wedge \left(x=A\wedge \phi \right)\right)$$
4	3	exbii	$$⊢\exists x\left(x=A\wedge \left(y=B\wedge \phi \right)\right)\leftrightarrow \exists x\left(y=B\wedge \left(x=A\wedge \phi \right)\right)$$
5		19.42v	$$⊢\exists x\left(y=B\wedge \left(x=A\wedge \phi \right)\right)\leftrightarrow \left(y=B\wedge \exists x\left(x=A\wedge \phi \right)\right)$$
6	4 5	bitri	$$⊢\exists x\left(x=A\wedge \left(y=B\wedge \phi \right)\right)\leftrightarrow \left(y=B\wedge \exists x\left(x=A\wedge \phi \right)\right)$$
7	6	exbii	$$⊢\exists y\exists x\left(x=A\wedge \left(y=B\wedge \phi \right)\right)\leftrightarrow \exists y\left(y=B\wedge \exists x\left(x=A\wedge \phi \right)\right)$$
8	1 2 7	3bitr3i	$$⊢\exists x\left(x=A\wedge \exists y\left(y=B\wedge \phi \right)\right)\leftrightarrow \exists y\left(y=B\wedge \exists x\left(x=A\wedge \phi \right)\right)$$
9		sbc5	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\exists y\left(y=B\wedge \phi \right)\leftrightarrow \exists x\left(x=A\wedge \exists y\left(y=B\wedge \phi \right)\right)$$
10		sbc5	$$⊢\underset{\dot{}}{[}B/y\underset{\dot{}}{]}\exists x\left(x=A\wedge \phi \right)\leftrightarrow \exists y\left(y=B\wedge \exists x\left(x=A\wedge \phi \right)\right)$$
11	8 9 10	3bitr4i	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\exists y\left(y=B\wedge \phi \right)\leftrightarrow \underset{\dot{}}{[}B/y\underset{\dot{}}{]}\exists x\left(x=A\wedge \phi \right)$$
12		sbc5	$$⊢\underset{\dot{}}{[}B/y\underset{\dot{}}{]}\phi \leftrightarrow \exists y\left(y=B\wedge \phi \right)$$
13	12	sbcbii	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\underset{\dot{}}{[}B/y\underset{\dot{}}{]}\phi \leftrightarrow \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\exists y\left(y=B\wedge \phi \right)$$
14		sbc5	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \leftrightarrow \exists x\left(x=A\wedge \phi \right)$$
15	14	sbcbii	$$⊢\underset{\dot{}}{[}B/y\underset{\dot{}}{]}\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \leftrightarrow \underset{\dot{}}{[}B/y\underset{\dot{}}{]}\exists x\left(x=A\wedge \phi \right)$$
16	11 13 15	3bitr4i	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\underset{\dot{}}{[}B/y\underset{\dot{}}{]}\phi \leftrightarrow \underset{\dot{}}{[}B/y\underset{\dot{}}{]}\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi$$