Theorem sbcbi2

Description: Substituting into equivalent wff's gives equivalent results. (Contributed by Giovanni Mascellani, 9-Apr-2018) (Proof shortened by Wolf Lammen, 4-May-2023) Avoid ax-10, ax-12. (Revised by Steven Nguyen, 5-May-2024)

	Ref	Expression
Assertion	sbcbi2	$$⊢\forall x\left(\phi \leftrightarrow \psi \right)\to \left(\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \leftrightarrow \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\psi \right)$$

Proof

Step	Hyp	Ref	Expression
1		abbi	$$⊢\forall x\left(\phi \leftrightarrow \psi \right)\to \left\{x|\phi \right\}=\left\{x|\psi \right\}$$
2		eleq2	$$⊢\left\{x|\phi \right\}=\left\{x|\psi \right\}\to \left(A\in \left\{x|\phi \right\}\leftrightarrow A\in \left\{x|\psi \right\}\right)$$
3	1 2	syl	$$⊢\forall x\left(\phi \leftrightarrow \psi \right)\to \left(A\in \left\{x|\phi \right\}\leftrightarrow A\in \left\{x|\psi \right\}\right)$$
4		df-sbc	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \leftrightarrow A\in \left\{x|\phi \right\}$$
5		df-sbc	$$⊢\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\psi \leftrightarrow A\in \left\{x|\psi \right\}$$
6	3 4 5	3bitr4g	$$⊢\forall x\left(\phi \leftrightarrow \psi \right)\to \left(\underset{\dot{}}{[}A/x\underset{\dot{}}{]}\phi \leftrightarrow \underset{\dot{}}{[}A/x\underset{\dot{}}{]}\psi \right)$$