Theorem unopab 5185

Description: Union of two ordered pair class abstractions. (Contributed by NM, 30-Sep-2002.)

Assertion

Ref	Expression
unopab	⊢ ({⟨𝑥, 𝑦⟩ ∣ 𝜑} ∪ {⟨𝑥, 𝑦⟩ ∣ 𝜓}) = {⟨𝑥, 𝑦⟩ ∣ (𝜑 ∨ 𝜓)}

Proof of Theorem unopab

Dummy variables 𝑧 𝑤 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		eqeq1 2765	. . . . . 6 ⊢ (𝑧 = 𝑤 → (𝑧 = ⟨𝑥, 𝑦⟩ ↔ 𝑤 = ⟨𝑥, 𝑦⟩))
2	1	anbi1d 643	. . . . 5 ⊢ (𝑧 = 𝑤 → ((𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ↔ (𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜑)))
3	2	2exbidv 1957	. . . 4 ⊢ (𝑧 = 𝑤 → (∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ↔ ∃𝑥∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜑)))
4	1	anbi1d 643	. . . . 5 ⊢ (𝑧 = 𝑤 → ((𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓) ↔ (𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)))
5	4	2exbidv 1957	. . . 4 ⊢ (𝑧 = 𝑤 → (∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓) ↔ ∃𝑥∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)))
6	3, 5	unabw 4253	. . 3 ⊢ ({𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑)} ∪ {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)}) = {𝑤 ∣ (∃𝑥∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ∨ ∃𝑥∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜓))}
7		19.43 1915	. . . . 5 ⊢ (∃𝑥(∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ∨ ∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)) ↔ (∃𝑥∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ∨ ∃𝑥∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)))
8		andi 1025	. . . . . . . 8 ⊢ ((𝑤 = ⟨𝑥, 𝑦⟩ ∧ (𝜑 ∨ 𝜓)) ↔ ((𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ∨ (𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)))
9	8	exbii 1881	. . . . . . 7 ⊢ (∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ (𝜑 ∨ 𝜓)) ↔ ∃𝑦((𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ∨ (𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)))
10		19.43 1915	. . . . . . 7 ⊢ (∃𝑦((𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ∨ (𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)) ↔ (∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ∨ ∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)))
11	9, 10	bitr2i 279	. . . . . 6 ⊢ ((∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ∨ ∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)) ↔ ∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ (𝜑 ∨ 𝜓)))
12	11	exbii 1881	. . . . 5 ⊢ (∃𝑥(∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ∨ ∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)) ↔ ∃𝑥∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ (𝜑 ∨ 𝜓)))
13	7, 12	bitr3i 280	. . . 4 ⊢ ((∃𝑥∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ∨ ∃𝑥∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)) ↔ ∃𝑥∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ (𝜑 ∨ 𝜓)))
14	13	abbii 2828	. . 3 ⊢ {𝑤 ∣ (∃𝑥∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) ∨ ∃𝑥∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ 𝜓))} = {𝑤 ∣ ∃𝑥∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ (𝜑 ∨ 𝜓))}
15	6, 14	eqtri 2784	. 2 ⊢ ({𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑)} ∪ {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)}) = {𝑤 ∣ ∃𝑥∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ (𝜑 ∨ 𝜓))}
16		df-opab 5168	. . 3 ⊢ {⟨𝑥, 𝑦⟩ ∣ 𝜑} = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑)}
17		df-opab 5168	. . 3 ⊢ {⟨𝑥, 𝑦⟩ ∣ 𝜓} = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)}
18	16, 17	uneq12i 4113	. 2 ⊢ ({⟨𝑥, 𝑦⟩ ∣ 𝜑} ∪ {⟨𝑥, 𝑦⟩ ∣ 𝜓}) = ({𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑)} ∪ {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)})
19		df-opab 5168	. 2 ⊢ {⟨𝑥, 𝑦⟩ ∣ (𝜑 ∨ 𝜓)} = {𝑤 ∣ ∃𝑥∃𝑦(𝑤 = ⟨𝑥, 𝑦⟩ ∧ (𝜑 ∨ 𝜓))}
20	15, 18, 19	3eqtr4i 2794	1 ⊢ ({⟨𝑥, 𝑦⟩ ∣ 𝜑} ∪ {⟨𝑥, 𝑦⟩ ∣ 𝜓}) = {⟨𝑥, 𝑦⟩ ∣ (𝜑 ∨ 𝜓)}

This proof depends on syntax axioms:   ∧ wa 401   ∨ wo 861   = wceq 1570  ∃wex 1812  {cab 2739   ∪ cun 3897  ⟨cop 4590  {copab 5167

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-8 2147  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-v 3453  df-un 3904  df-opab 5168

This theorem is used by:  xpundi  5720  xpundir  5721  cnvun  6131  coundi  6241  coundir  6242  mptun  6677  opsrtoslem1  22344  lgsquadlem3  27691