Theorem bnj956 34769

Description: First-order logic and set theory. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.) (New usage is discouraged.)

Hypothesis

Ref	Expression
bnj956.1	⊢ (𝐴 = 𝐵 → ∀𝑥 𝐴 = 𝐵)

Assertion

Ref	Expression
bnj956	⊢ (𝐴 = 𝐵 → ∪ 𝑥 ∈ 𝐴 𝐶 = ∪ 𝑥 ∈ 𝐵 𝐶)

Proof of Theorem bnj956

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		bnj956.1	. . . 4 ⊢ (𝐴 = 𝐵 → ∀𝑥 𝐴 = 𝐵)
2		eleq2 2828	. . . . . . 7 ⊢ (𝐴 = 𝐵 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
3	2	anbi1d 631	. . . . . 6 ⊢ (𝐴 = 𝐵 → ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)))
4	3	alexbii 1830	. . . . 5 ⊢ (∀𝑥 𝐴 = 𝐵 → (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶)))
5		df-rex 3069	. . . . 5 ⊢ (∃𝑥 ∈ 𝐴 𝑦 ∈ 𝐶 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐶))
6		df-rex 3069	. . . . 5 ⊢ (∃𝑥 ∈ 𝐵 𝑦 ∈ 𝐶 ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶))
7	4, 5, 6	3bitr4g 314	. . . 4 ⊢ (∀𝑥 𝐴 = 𝐵 → (∃𝑥 ∈ 𝐴 𝑦 ∈ 𝐶 ↔ ∃𝑥 ∈ 𝐵 𝑦 ∈ 𝐶))
8	1, 7	syl 17	. . 3 ⊢ (𝐴 = 𝐵 → (∃𝑥 ∈ 𝐴 𝑦 ∈ 𝐶 ↔ ∃𝑥 ∈ 𝐵 𝑦 ∈ 𝐶))
9	8	abbidv 2806	. 2 ⊢ (𝐴 = 𝐵 → {𝑦 ∣ ∃𝑥 ∈ 𝐴 𝑦 ∈ 𝐶} = {𝑦 ∣ ∃𝑥 ∈ 𝐵 𝑦 ∈ 𝐶})
10		df-iun 4998	. 2 ⊢ ∪ 𝑥 ∈ 𝐴 𝐶 = {𝑦 ∣ ∃𝑥 ∈ 𝐴 𝑦 ∈ 𝐶}
11		df-iun 4998	. 2 ⊢ ∪ 𝑥 ∈ 𝐵 𝐶 = {𝑦 ∣ ∃𝑥 ∈ 𝐵 𝑦 ∈ 𝐶}
12	9, 10, 11	3eqtr4g 2800	1 ⊢ (𝐴 = 𝐵 → ∪ 𝑥 ∈ 𝐴 𝐶 = ∪ 𝑥 ∈ 𝐵 𝐶)

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395  ∀wal 1535   = wceq 1537  ∃wex 1776   ∈ wcel 2106  {cab 2712  ∃wrex 3068  ∪ ciun 4996

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-rex 3069  df-iun 4998

This theorem is referenced by:  bnj1316  34813  bnj953  34932  bnj1000  34934  bnj966  34937