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Metamath Proof Explorer

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	$⊢ φ$
	Assertion	sbtlem	$⊢ \forall y (y = t \to \forall x (x = y \to φ))$

Proof

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