Theorem sbnf

Description: Move nonfree predicate in and out of substitution; see sbal and sbex . (Contributed by BJ, 2-May-2019) (Proof shortened by Wolf Lammen, 2-May-2025)

	Ref	Expression
Assertion	sbnf	$$⊢\left[z/y\right]Ⅎx\phi \leftrightarrow Ⅎx\left[z/y\right]\phi$$

Proof

Step	Hyp	Ref	Expression
1		df-nf	$$⊢Ⅎx\phi \leftrightarrow \left(\exists x\phi \to \forall x\phi \right)$$
2	1	sbbii	$$⊢\left[z/y\right]Ⅎx\phi \leftrightarrow \left[z/y\right]\left(\exists x\phi \to \forall x\phi \right)$$
3		sbim	$$⊢\left[z/y\right]\left(\exists x\phi \to \forall x\phi \right)\leftrightarrow \left(\left[z/y\right]\exists x\phi \to \left[z/y\right]\forall x\phi \right)$$
4		sbex	$$⊢\left[z/y\right]\exists x\phi \leftrightarrow \exists x\left[z/y\right]\phi$$
5		sbal	$$⊢\left[z/y\right]\forall x\phi \leftrightarrow \forall x\left[z/y\right]\phi$$
6	4 5	imbi12i	$$⊢\left(\left[z/y\right]\exists x\phi \to \left[z/y\right]\forall x\phi \right)\leftrightarrow \left(\exists x\left[z/y\right]\phi \to \forall x\left[z/y\right]\phi \right)$$
7		df-nf	$$⊢Ⅎx\left[z/y\right]\phi \leftrightarrow \left(\exists x\left[z/y\right]\phi \to \forall x\left[z/y\right]\phi \right)$$
8	6 7	bitr4i	$$⊢\left(\left[z/y\right]\exists x\phi \to \left[z/y\right]\forall x\phi \right)\leftrightarrow Ⅎx\left[z/y\right]\phi$$
9	2 3 8	3bitri	$$⊢\left[z/y\right]Ⅎx\phi \leftrightarrow Ⅎx\left[z/y\right]\phi$$