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Metamath Proof Explorer

Theorem sb6x

Description: Equivalence involving substitution for a variable not free. Usage of this theorem is discouraged because it depends on ax-13 . Usage of sb6 is preferred, which requires fewer axioms. (Contributed by NM, 2-Jun-1993) (Revised by Mario Carneiro, 4-Oct-2016) (New usage is discouraged.)

		Ref	Expression
	Hypothesis	sb6x.1	⊢ Ⅎ 𝑥 𝜑
	Assertion	sb6x	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) )

Proof

Step	Hyp	Ref	Expression
1		sb6x.1	⊢ Ⅎ 𝑥 𝜑
2	1	sbf	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ 𝜑 )
3		biidd	⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜑 ) )
4	1 3	equsal	⊢ ( ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) ↔ 𝜑 )
5	2 4	bitr4i	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) )