Theorem unab 3402

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 1954	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
2		df-clab 2164	. . 3 ⊢ (𝑦 ∈ {𝑥 ∣ (𝜑 ∨ 𝜓)} ↔ [𝑦 / 𝑥](𝜑 ∨ 𝜓))
3		df-clab 2164	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
4		df-clab 2164	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓)
5	3, 4	orbi12i 764	. . 3 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∨ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
6	1, 2, 5	3bitr4ri 213	. 2 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ∨ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ 𝑦 ∈ {𝑥 ∣ (𝜑 ∨ 𝜓)})
7	6	uneqri 3277	1 ⊢ ({𝑥 ∣ 𝜑} ∪ {𝑥 ∣ 𝜓}) = {𝑥 ∣ (𝜑 ∨ 𝜓)}

Syntax hints:   ∨ wo 708   = wceq 1353  [wsb 1762   ∈ wcel 2148  {cab 2163   ∪ cun 3127

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2739  df-un 3133

This theorem is referenced by:  unrab  3406  rabun2  3414  dfif6  3536  unopab  4082  dmun  4834  frecabex  6398