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Metamath Proof Explorer

Theorem sbcom3vv

Description: Substituting y for x and then z for y is equivalent to substituting z for both x and y . Version of sbcom3 with a disjoint variable condition using fewer axioms. (Contributed by NM, 27-May-1997) (Revised by Giovanni Mascellani, 8-Apr-2018) (Revised by BJ, 30-Dec-2020) (Proof shortened by Wolf Lammen, 19-Jan-2023)

		Ref	Expression
	Assertion	sbcom3vv	\|- ( [ z / y ] [ y / x ] ph <-> [ z / y ] [ z / x ] ph )

Proof

Step	Hyp	Ref	Expression
1		sbequ	\|- ( y = z -> ( [ y / x ] ph <-> [ z / x ] ph ) )
2	1	sbbiiev	\|- ( [ z / y ] [ y / x ] ph <-> [ z / y ] [ z / x ] ph )