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Metamath Proof Explorer

Theorem hbsbw

Description: If z is not free in ph , it is not free in [ y / x ] ph when y and z are distinct. Version of hbsb with a disjoint variable condition, which requires fewer axioms. (Contributed by NM, 12-Aug-1993) Remove dependencies on axioms. (Revised by GG, 23-May-2024) (Proof shortened by Wolf Lammen, 14-May-2025)

		Ref	Expression
	Hypothesis	hbsbw.1	⊢ ( 𝜑 → ∀ 𝑧 𝜑 )
	Assertion	hbsbw	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 → ∀ 𝑧 [ 𝑦 / 𝑥 ] 𝜑 )

Proof

Step	Hyp	Ref	Expression
1		hbsbw.1	⊢ ( 𝜑 → ∀ 𝑧 𝜑 )
2	1	sbimi	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 → [ 𝑦 / 𝑥 ] ∀ 𝑧 𝜑 )
3		sbal	⊢ ( [ 𝑦 / 𝑥 ] ∀ 𝑧 𝜑 ↔ ∀ 𝑧 [ 𝑦 / 𝑥 ] 𝜑 )
4	2 3	sylib	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 → ∀ 𝑧 [ 𝑦 / 𝑥 ] 𝜑 )