Theorem parteq2 39953

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 2775	. . 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 39947	. 2 ⊢ (𝑅 Part 𝐴 ↔ ( Disj 𝑅 ∧ (dom 𝑅 / 𝑅) = 𝐴))
4		dfpart2 39947	. 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 1569  dom cdm 5655   / cqs 8730   Disj wdisjALTV 39294   Part wpart 39299

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-cleq 2755  df-dmqs 39798  df-part 39944

This theorem is used by:  parteq12  39954