Theorem bj-unrab 37408

Description: Generalization of unrab 4267. 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 4130	. . . 4 ⊢ 𝐴 ⊆ (𝐴 ∪ 𝐵)
2		rabss2 4030	. . . 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 878	. . . . 5 ⊢ (𝜑 → (𝜑 ∨ 𝜓))
5	4	a1i 11	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) → (𝜑 → (𝜑 ∨ 𝜓)))
6	5	ss2rabi 4029	. . 3 ⊢ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ 𝜑} ⊆ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ (𝜑 ∨ 𝜓)}
7	3, 6	sstri 3945	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} ⊆ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ (𝜑 ∨ 𝜓)}
8		ssun2 4131	. . . 4 ⊢ 𝐵 ⊆ (𝐴 ∪ 𝐵)
9		rabss2 4030	. . . 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 879	. . . . 5 ⊢ (𝜓 → (𝜑 ∨ 𝜓))
12	11	a1i 11	. . . 4 ⊢ (𝑥 ∈ (𝐴 ∪ 𝐵) → (𝜓 → (𝜑 ∨ 𝜓)))
13	12	ss2rabi 4029	. . 3 ⊢ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ 𝜓} ⊆ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ (𝜑 ∨ 𝜓)}
14	10, 13	sstri 3945	. 2 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜓} ⊆ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ (𝜑 ∨ 𝜓)}
15	7, 14	unssi 4143	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐵 ∣ 𝜓}) ⊆ {𝑥 ∈ (𝐴 ∪ 𝐵) ∣ (𝜑 ∨ 𝜓)}

Syntax hints:   → wi 4   ∨ wo 858   ∈ wcel 2142  {crab 3414   ∪ cun 3902   ⊆ wss 3904

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-ext 2734

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1563  df-ex 1800  df-sb 2091  df-clab 2741  df-cleq 2754  df-clel 2837  df-ral 3077  df-rab 3415  df-v 3456  df-un 3909  df-ss 3921

This theorem is referenced by: (None)