Theorem bj-unrab 38054

Description: Generalization of unrab 4263. Equality need not hold. (Contributed by BJ, 21-Apr-2019.)

Assertion

Ref	Expression
bj-unrab	⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐵 ∣ 𝜓}) ⊆ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ (𝜑 ∨ 𝜓)}

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑥)

Proof of Theorem bj-unrab

Step	Hyp	Ref	Expression
1		ssun1 4179	. . . 4 ⊢ 𝐴 ⊆ (𝐴 ∪ 𝐵)
2		rabss2 4005	. . . 4 ⊢ (𝐴 ⊆ (𝐴 ∪ 𝐵) → {𝑥 ∈ 𝐴 ∣ 𝜑} ⊆ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ 𝜑})
3	1, 2	ax-mp 5	. . 3 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} ⊆ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ 𝜑}
4		orc 881	. . . . 5 ⊢ (𝜑 → (𝜑 ∨ 𝜓))
5	4	a1i 11	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) → (𝜑 → (𝜑 ∨ 𝜓)))
6	5	ss2rabi 4004	. . 3 ⊢ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ 𝜑} ⊆ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ (𝜑 ∨ 𝜓)}
7	3, 6	sstri 3922	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} ⊆ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ (𝜑 ∨ 𝜓)}
8		ssun2 4180	. . . 4 ⊢ 𝐵 ⊆ (𝐴 ∪ 𝐵)
9		rabss2 4005	. . . 4 ⊢ (𝐵 ⊆ (𝐴 ∪ 𝐵) → {𝑥 ∈ 𝐵 ∣ 𝜓} ⊆ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ 𝜓})
10	8, 9	ax-mp 5	. . 3 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜓} ⊆ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ 𝜓}
11		olc 882	. . . . 5 ⊢ (𝜓 → (𝜑 ∨ 𝜓))
12	11	a1i 11	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) → (𝜓 → (𝜑 ∨ 𝜓)))
13	12	ss2rabi 4004	. . 3 ⊢ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ 𝜓} ⊆ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ (𝜑 ∨ 𝜓)}
14	10, 13	sstri 3922	. 2 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜓} ⊆ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ (𝜑 ∨ 𝜓)}
15	7, 14	unssi 4192	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐵 ∣ 𝜓}) ⊆ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ (𝜑 ∨ 𝜓)}

This proof depends on syntax axioms:   → wi 4   ∨ wo 861   ∈ wcel 2147  {crab 3416   ⊆ wss 3900   ∪ cun 4152

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rab 3417  df-v 3456  df-ss 3901  df-un 4154

This theorem is used by: (None)