Theorem unabw 4259

Description: Union of two class abstractions. Version of unab 4260 using implicit substitution, which does not require ax-8 2143, ax-10 2174, ax-12 2211. (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 3909	. 2 ⊢ ({𝑥 ∣ 𝜑} ∪ {𝑥 ∣ 𝜓}) = {𝑦 ∣ (𝑦 ∈ {𝑥 ∣ 𝜑} ∨ 𝑦 ∈ {𝑥 ∣ 𝜓})}
2		df-clab 2740	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
3		unabw.1	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜒))
4	3	sbievw 2126	. . . . 5 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜒)
5	2, 4	bitri 277	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜒)
6		df-clab 2740	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓)
7		unabw.2	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜃))
8	7	sbievw 2126	. . . . 5 ⊢ ([𝑦 / 𝑥]𝜓 ↔ 𝜃)
9	6, 8	bitri 277	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ 𝜃)
10	5, 9	orbi12i 925	. . 3 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∨ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ (𝜒 ∨ 𝜃))
11	10	abbii 2828	. 2 ⊢ {𝑦 ∣ (𝑦 ∈ {𝑥 ∣ 𝜑} ∨ 𝑦 ∈ {𝑥 ∣ 𝜓})} = {𝑦 ∣ (𝜒 ∨ 𝜃)}
12	1, 11	eqtri 2784	1 ⊢ ({𝑥 ∣ 𝜑} ∪ {𝑥 ∣ 𝜓}) = {𝑦 ∣ (𝜒 ∨ 𝜃)}

Syntax hints:   → wi 4   ↔ wb 208   ∨ wo 858   = wceq 1559  [wsb 2089   ∈ wcel 2141  {cab 2739   ∪ cun 3902

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-9 2151  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-ex 1799  df-sb 2090  df-clab 2740  df-cleq 2753  df-un 3909

This theorem is referenced by:  dfif6  4482  unopab  5179  dmun  5884