Theorem unab 4248

Description: Union of two class abstractions. (Contributed by NM, 29-Sep-2002.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)

Assertion

Ref	Expression
unab	⊢ ({𝑥 ∣ 𝜑} ∪ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∨ 𝜓)}

Proof of Theorem unab

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sbor 2313	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
2		df-clab 2715	. . 3 ⊢ (𝑦 ∈ {𝑥 ∣ (𝜑 ∨ 𝜓)} ↔ [𝑦 / 𝑥](𝜑 ∨ 𝜓))
3		df-clab 2715	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
4		df-clab 2715	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓)
5	3, 4	orbi12i 915	. . 3 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∨ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
6	1, 2, 5	3bitr4ri 304	. 2 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∨ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ 𝑦 ∈ {𝑥 ∣ (𝜑 ∨ 𝜓)})
7	6	uneqri 4096	1 ⊢ ({𝑥 ∣ 𝜑} ∪ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∨ 𝜓)}

Syntax hints:   ∨ wo 848   = wceq 1542  [wsb 2068   ∈ wcel 2114  {cab 2714   ∪ cun 3887

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-10 2147  ax-12 2185  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-ex 1782  df-nf 1786  df-sb 2069  df-clab 2715  df-cleq 2728  df-clel 2811  df-v 3431  df-un 3894

This theorem is referenced by:  unrab  4255  rabun2  4264  hashf1lem2  14418  vdwlem6  16957  addsasslem1  27995  addsasslem2  27996  addsdilem1  28143  addsdilem2  28144  mulsasslem1  28155  mulsasslem2  28156  vtxdun  29550  satfvsuclem1  35541  satf0suclem  35557  fmlasuc0  35566  ecun  38714  sticksstones22  42607  diophun  43205