Metamath Proof Explorer


Theorem parteq1

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

Ref Expression
Assertion parteq1 ⊢ R = S → R Part A ↔ S Part A

Proof

Step Hyp Ref Expression
1 disjdmqseqeq1 ⊢ R = S → Disj R ∧ dom ⁡ R / R = A ↔ Disj S ∧ dom ⁡ S / S = A
2 dfpart2 ⊢ R Part A ↔ Disj R ∧ dom ⁡ R / R = A
3 dfpart2 ⊢ S Part A ↔ Disj S ∧ dom ⁡ S / S = A
4 1 2 3 3bitr4g ⊢ R = S → R Part A ↔ S Part A