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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	$⊢ φ \to \forall z φ$
	Assertion	hbsbw	$⊢ [y/ x] φ \to \forall z [y/ x] φ$

Proof

Step	Hyp	Ref	Expression
1		hbsbw.1	$⊢ φ \to \forall z φ$
2	1	sbimi	$⊢ [y/ x] φ \to [y/ x] \forall z φ$
3		sbal	$⊢ [y/ x] \forall z φ \leftrightarrow \forall z [y/ x] φ$
4	2 3	sylib	$⊢ [y/ x] φ \to \forall z [y/ x] φ$