Metamath Proof Explorer


Theorem parteq1i

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

Ref Expression
Hypothesis parteq1i.1 ⊢ R = S
Assertion parteq1i ⊢ R Part A ↔ S Part A

Proof

Step Hyp Ref Expression
1 parteq1i.1 ⊢ R = S
2 parteq1 ⊢ R = S → R Part A ↔ S Part A
3 1 2 ax-mp ⊢ R Part A ↔ S Part A