Theorem unabw 4260

Description: Union of two class abstractions. Version of unab 4261 using implicit substitution, which does not require ax-8 2145, ax-10 2176, ax-12 2213. (Contributed by GG, 15-Oct-2024.)

Hypotheses

Ref	Expression
unabw.1	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜒))
unabw.2	⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜃))

Assertion

Ref	Expression
unabw	⊢ ({𝑥 ∣ 𝜑} ∪ {𝑥 ∣ 𝜓}) = {𝑦 ∣ (𝜒 ∨ 𝜃)}

Distinct variable groups:   𝑥,𝑦   𝜑,𝑦   𝜓,𝑦   𝜒,𝑥   𝜃,𝑥

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑥)   𝜒(𝑦)   𝜃(𝑦)

Proof of Theorem unabw

Step	Hyp	Ref	Expression
1		df-un 3910	. 2 ⊢ ({𝑥 ∣ 𝜑} ∪ {𝑥 ∣ 𝜓}) = {𝑦 ∣ (𝑦 ∈ {𝑥 ∣ 𝜑} ∨ 𝑦 ∈ {𝑥 ∣ 𝜓})}
2		df-clab 2742	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
3		unabw.1	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜒))
4	3	sbievw 2128	. . . . 5 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜒)
5	2, 4	bitri 278	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜒)
6		df-clab 2742	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓)
7		unabw.2	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜃))
8	7	sbievw 2128	. . . . 5 ⊢ ([𝑦 / 𝑥]𝜓 ↔ 𝜃)
9	6, 8	bitri 278	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ 𝜃)
10	5, 9	orbi12i 927	. . 3 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∨ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ (𝜒 ∨ 𝜃))
11	10	abbii 2830	. 2 ⊢ {𝑦 ∣ (𝑦 ∈ {𝑥 ∣ 𝜑} ∨ 𝑦 ∈ {𝑥 ∣ 𝜓})} = {𝑦 ∣ (𝜒 ∨ 𝜃)}
12	1, 11	eqtri 2786	1 ⊢ ({𝑥 ∣ 𝜑} ∪ {𝑥 ∣ 𝜓}) = {𝑦 ∣ (𝜒 ∨ 𝜃)}

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∨ wo 860   = wceq 1570  [wsb 2096   ∈ wcel 2143  {cab 2741   ∪ cun 3903

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-9 2153  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-un 3910

This theorem is used by:  dfif6  4490  unopab  5191  dmun  5900