Theorem unab 4253

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 2339	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
2		df-clab 2739	. . 3 ⊢ (𝑦 ∈ {𝑥 ∣ (𝜑 ∨ 𝜓)} ↔ [𝑦 / 𝑥](𝜑 ∨ 𝜓))
3		df-clab 2739	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
4		df-clab 2739	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓)
5	3, 4	orbi12i 928	. . 3 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∨ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
6	1, 2, 5	3bitr4ri 307	. 2 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∨ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ 𝑦 ∈ {𝑥 ∣ (𝜑 ∨ 𝜓)})
7	6	uneqri 4155	1 ⊢ ({𝑥 ∣ 𝜑} ∪ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∨ 𝜓)}

This proof depends on syntax axioms:   ∨ wo 861   = wceq 1569  [wsb 2098   ∈ wcel 2144  {cab 2738   ∪ cun 4149

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-un 4151

This theorem is used by:  unrab  4260  rabun2  4269  hashf1lem2  14605  vdwlem6  17170  addsasslem1  28395  addsasslem2  28396  addsdilem1  28543  addsdilem2  28544  mulsasslem1  28555  mulsasslem2  28556  vtxdun  30068  satfvsuclem1  36211  satf0suclem  36227  fmlasuc0  36236  dfproplem  38735  ecun  39419  sticksstones22  43312  diophun  43878