Theorem unab 3338

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 1925	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
2		df-clab 2124	. . 3 ⊢ (𝑦 ∈ {𝑥 ∣ (𝜑 ∨ 𝜓)} ↔ [𝑦 / 𝑥](𝜑 ∨ 𝜓))
3		df-clab 2124	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
4		df-clab 2124	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓)
5	3, 4	orbi12i 753	. . 3 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∨ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
6	1, 2, 5	3bitr4ri 212	. 2 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∨ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ 𝑦 ∈ {𝑥 ∣ (𝜑 ∨ 𝜓)})
7	6	uneqri 3213	1 ⊢ ({𝑥 ∣ 𝜑} ∪ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∨ 𝜓)}

Syntax hints:   ∨ wo 697   = wceq 1331   ∈ wcel 1480  [wsb 1735  {cab 2123   ∪ cun 3064

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2119

This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2124  df-cleq 2130  df-clel 2133  df-nfc 2268  df-v 2683  df-un 3070

This theorem is referenced by:  unrab  3342  rabun2  3350  dfif6  3471  unopab  4002  dmun  4741  frecabex  6288