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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	⊢ ∀ 𝑦 ( 𝑦 = 𝑡 → ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) )

Proof

Step	Hyp	Ref	Expression
1		sbtlem.1	⊢ 𝜑
2	1	a1i	⊢ ( 𝑥 = 𝑦 → 𝜑 )
3	2	ax-gen	⊢ ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 )
4	3	a1i	⊢ ( 𝑦 = 𝑡 → ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) )
5	4	ax-gen	⊢ ∀ 𝑦 ( 𝑦 = 𝑡 → ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) )