Theorem sbtlem

Description: In the case of sbt , the hypothesis in df-sb is derivable from propositional axioms and ax-gen alone. The essential proof step is presented in this lemma. (Contributed by Wolf Lammen, 4-Feb-2026)

	Ref	Expression
Hypothesis	sbtlem.1	$$⊢\phi$$
Assertion	sbtlem	$$⊢\forall y\left(y=t\to \forall x\left(x=y\to \phi \right)\right)$$

Proof

Step	Hyp	Ref	Expression
1		sbtlem.1	$$⊢\phi$$
2	1	a1i	$$⊢x=y\to \phi$$
3	2	ax-gen	$$⊢\forall x\left(x=y\to \phi \right)$$
4	3	a1i	$$⊢y=t\to \forall x\left(x=y\to \phi \right)$$
5	4	ax-gen	$$⊢\forall y\left(y=t\to \forall x\left(x=y\to \phi \right)\right)$$