Theorem parteq1d

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

	Ref	Expression
Hypothesis	parteq1d.1	⊢ ( 𝜑 → 𝑅 = 𝑆 )
Assertion	parteq1d	⊢ ( 𝜑 → ( 𝑅 Part 𝐴 ↔ 𝑆 Part 𝐴 ) )

Proof

Step	Hyp	Ref	Expression
1		parteq1d.1	⊢ ( 𝜑 → 𝑅 = 𝑆 )
2		parteq1	⊢ ( 𝑅 = 𝑆 → ( 𝑅 Part 𝐴 ↔ 𝑆 Part 𝐴 ) )
3	1 2	syl	⊢ ( 𝜑 → ( 𝑅 Part 𝐴 ↔ 𝑆 Part 𝐴 ) )