Theorem unab 1691

Description: Union of two class abstractions.

Assertion

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

Proof of Theorem unab

Step	Hyp	Ref	Expression
1		df-clab 1093	. . . . 5 ⊢ (y ∈ {x∣φ} ↔ [y / x]φ)
2		df-clab 1093	. . . . 5 ⊢ (y ∈ {x∣ψ} ↔ [y / x]ψ)
3	1, 2	orbi12i 216	. . . 4 ⊢ ((y ∈ {x∣φ} ∨ y ∈ {x∣ψ}) ↔ ([y / x]φ ∨ [y / x]ψ))
4		sbor 887	. . . 4 ⊢ ([y / x](φ ∨ ψ) ↔ ([y / x]φ ∨ [y / x]ψ))
5	3, 4	bitr4 154	. . 3 ⊢ ((y ∈ {x∣φ} ∨ y ∈ {x∣ψ}) ↔ [y / x](φ ∨ ψ))
6		elun 1601	. . 3 ⊢ (y ∈ ({x∣φ} ∪ {x∣ψ}) ↔ (y ∈ {x∣φ} ∨ y ∈ {x∣ψ}))
7		df-clab 1093	. . 3 ⊢ (y ∈ {x∣(φ ∨ ψ)} ↔ [y / x](φ ∨ ψ))
8	5, 6, 7	3bitr4 158	. 2 ⊢ (y ∈ ({x∣φ} ∪ {x∣ψ}) ↔ y ∈ {x∣(φ ∨ ψ)})
9	8	cleqri 1101	1 ⊢ ({x∣φ} ∪ {x∣ψ}) = {x∣(φ ∨ ψ)}

Syntax hints:   ∨ wo 195  [wsb 852  {cab 1090   = wceq 1091   ∈ wcel 1092   ∪ cun 1485

This theorem is referenced by:  unrab 1694  unopab 2121  infxpidmlem9 4941

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-un 1490

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