Theorem 3unrab 33147

Description: Union of three restricted class abstractions. (Contributed by Thierry Arnoux, 6-Jul-2025.)

Assertion

Ref	Expression
3unrab	⊢ (({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) ∪ {𝑥 ∈ 𝐴 ∣ 𝜒}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓 ∨ 𝜒)}

Proof of Theorem 3unrab

Step	Hyp	Ref	Expression
1		unrab 4260	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} ∪ {𝑥 ∈ 𝐴 ∣ 𝜒}) = {𝑥 ∈ 𝐴 ∣ ((𝜑 ∨ 𝜓) ∨ 𝜒)}
2		unrab 4260	. . 3 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)}
3	2	uneq1i 4163	. 2 ⊢ (({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) ∪ {𝑥 ∈ 𝐴 ∣ 𝜒}) = ({𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓)} ∪ {𝑥 ∈ 𝐴 ∣ 𝜒})
4		df-3or 1104	. . 3 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ((𝜑 ∨ 𝜓) ∨ 𝜒))
5	4	rabbii 3417	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓 ∨ 𝜒)} = {𝑥 ∈ 𝐴 ∣ ((𝜑 ∨ 𝜓) ∨ 𝜒)}
6	1, 3, 5	3eqtr4i 2793	1 ⊢ (({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐴 ∣ 𝜓}) ∪ {𝑥 ∈ 𝐴 ∣ 𝜒}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∨ 𝜓 ∨ 𝜒)}

This proof depends on syntax axioms:   ∨ wo 861   ∨ w3o 1102   = wceq 1569  {crab 3412   ∪ 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-3or 1104  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-un 4151

This theorem is used by:  constrfin  34425