Theorem parteq2 39587

Description: Equality theorem for partition. (Contributed by Peter Mazsa, 25-Jul-2024.)

Assertion

Ref	Expression
parteq2	⊢ (𝐴 = 𝐵 → (𝑅 Part 𝐴 ↔ 𝑅 Part 𝐵))

Proof of Theorem parteq2

Step	Hyp	Ref	Expression
1		eqeq2 2777	. . 3 ⊢ (𝐴 = 𝐵 → ((dom 𝑅 / 𝑅) = 𝐴 ↔ (dom 𝑅 / 𝑅) = 𝐵))
2	1	anbi2d 642	. 2 ⊢ (𝐴 = 𝐵 → (( Disj 𝑅 ∧ (dom 𝑅 / 𝑅) = 𝐴) ↔ ( Disj 𝑅 ∧ (dom 𝑅 / 𝑅) = 𝐵)))
3		dfpart2 39581	. 2 ⊢ (𝑅 Part 𝐴 ↔ ( Disj 𝑅 ∧ (dom 𝑅 / 𝑅) = 𝐴))
4		dfpart2 39581	. 2 ⊢ (𝑅 Part 𝐵 ↔ ( Disj 𝑅 ∧ (dom 𝑅 / 𝑅) = 𝐵))
5	2, 3, 4	3bitr4g 317	1 ⊢ (𝐴 = 𝐵 → (𝑅 Part 𝐴 ↔ 𝑅 Part 𝐵))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570  dom cdm 5663   / cqs 8699   Disj wdisjALTV 38928   Part wpart 38933

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-cleq 2757  df-dmqs 39432  df-part 39578

This theorem is used by:  parteq12  39588