Theorem parteq1d

Description: Equality theorem for partition, deduction version. (Contributed by Peter Mazsa, 5-Oct-2021)

	Ref	Expression
Hypothesis	parteq1d.1	|- ( ph -> R = S )
Assertion	parteq1d	|- ( ph -> ( R Part A <-> S Part A ) )

Proof

Step	Hyp	Ref	Expression
1		parteq1d.1	|- ( ph -> R = S )
2		parteq1	|- ( R = S -> ( R Part A <-> S Part A ) )
3	1 2	syl	|- ( ph -> ( R Part A <-> S Part A ) )