Theorem sbcco2

Description: A composition law for class substitution. Importantly, x may occur free in the class expression substituted for A . (Contributed by NM, 5-Sep-2004) (Proof shortened by Andrew Salmon, 8-Jun-2011)

	Ref	Expression
Hypothesis	sbcco2.1	$$⊢x=y\to A=B$$
Assertion	sbcco2	$$⊢\underset{\dot{}}{[}x/y\underset{\dot{}}{]}\underset{\dot{}}{[}B/x\underset{\dot{}}{]}\phi \leftrightarrow \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi$$

Proof

Step	Hyp	Ref	Expression
1		sbcco2.1	$$⊢x=y\to A=B$$
2		sbsbc	$$⊢\left[x/y\right]\underset{\dot{}}{[}B/x\underset{\dot{}}{]}\phi \leftrightarrow \underset{\dot{}}{[}x/y\underset{\dot{}}{]}\underset{\dot{}}{[}B/x\underset{\dot{}}{]}\phi$$
3	1	equcoms	$$⊢y=x\to A=B$$
4		dfsbcq	$$⊢A=B\to \left(\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \leftrightarrow \underset{\dot{}}{[}B/x\underset{\dot{}}{]}\phi \right)$$
5	4	bicomd	$$⊢A=B\to \left(\underset{\dot{}}{[}B/x\underset{\dot{}}{]}\phi \leftrightarrow \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \right)$$
6	3 5	syl	$$⊢y=x\to \left(\underset{\dot{}}{[}B/x\underset{\dot{}}{]}\phi \leftrightarrow \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \right)$$
7	6	sbievw	$$⊢\left[x/y\right]\underset{\dot{}}{[}B/x\underset{\dot{}}{]}\phi \leftrightarrow \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi$$
8	2 7	bitr3i	$$⊢\underset{\dot{}}{[}x/y\underset{\dot{}}{]}\underset{\dot{}}{[}B/x\underset{\dot{}}{]}\phi \leftrightarrow \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi$$