Theorem dfsb

Description: Simplify definition df-sb by proving the renaming independency. (Contributed by Wolf Lammen, 5-Feb-2026) df-sb changed. (Revised by Wolf Lammen, 4-Jun-2026)

	Ref	Expression
Assertion	dfsb	$$⊢\left[t/x\right]\phi \leftrightarrow \forall y\left(y=t\to \forall x\left(x=y\to \phi \right)\right)$$

Proof

Step	Hyp	Ref	Expression
1		dfsbimp	$$⊢\left[t/x\right]\phi \to \forall y\left(y=t\to \forall x\left(x=y\to \phi \right)\right)$$
2		df-sb	$$⊢\left[t/x\right]\phi \leftrightarrow \left(\forall y\left(y=t\to \forall x\left(x=y\to \phi \right)\right)\wedge \forall z\left(z=t\to \forall x\left(x=z\to \phi \right)\right)\right)$$
3		rename-sb	$$⊢\forall y\left(y=t\to \forall x\left(x=y\to \phi \right)\right)\leftrightarrow \forall z\left(z=t\to \forall x\left(x=z\to \phi \right)\right)$$
4	2 3	just3-df	$$⊢\forall y\left(y=t\to \forall x\left(x=y\to \phi \right)\right)\to \left[t/x\right]\phi$$
5	1 4	impbii	$$⊢\left[t/x\right]\phi \leftrightarrow \forall y\left(y=t\to \forall x\left(x=y\to \phi \right)\right)$$