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 ) )