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Metamath Proof Explorer

Theorem iunssOLD

Description: Obsolete version of iunss as of 2-Feb-2026. (Contributed by NM, 13-Sep-2003) (Proof shortened by Andrew Salmon, 25-Jul-2011) (Proof modification is discouraged.) (New usage is discouraged.)

		Ref	Expression
	Assertion	iunssOLD	⊢ ( ∪ 𝑥 ∈ 𝐴 𝐵 ⊆ 𝐶 ↔ ∀ 𝑥 ∈ 𝐴 𝐵 ⊆ 𝐶 )

Proof

Step	Hyp	Ref	Expression
1		df-iun	⊢ ∪ 𝑥 ∈ 𝐴 𝐵 = { 𝑦 ∣ ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 }
2	1	sseq1i	⊢ ( ∪ 𝑥 ∈ 𝐴 𝐵 ⊆ 𝐶 ↔ { 𝑦 ∣ ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 } ⊆ 𝐶 )
3		abss	⊢ ( { 𝑦 ∣ ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 } ⊆ 𝐶 ↔ ∀ 𝑦 ( ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) )
4		df-ss	⊢ ( 𝐵 ⊆ 𝐶 ↔ ∀ 𝑦 ( 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) )
5	4	ralbii	⊢ ( ∀ 𝑥 ∈ 𝐴 𝐵 ⊆ 𝐶 ↔ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ( 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) )
6		ralcom4	⊢ ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ( 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) ↔ ∀ 𝑦 ∀ 𝑥 ∈ 𝐴 ( 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) )
7		r19.23v	⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) ↔ ( ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) )
8	7	albii	⊢ ( ∀ 𝑦 ∀ 𝑥 ∈ 𝐴 ( 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) ↔ ∀ 𝑦 ( ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) )
9	5 6 8	3bitrri	⊢ ( ∀ 𝑦 ( ∃ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶 ) ↔ ∀ 𝑥 ∈ 𝐴 𝐵 ⊆ 𝐶 )
10	2 3 9	3bitri	⊢ ( ∪ 𝑥 ∈ 𝐴 𝐵 ⊆ 𝐶 ↔ ∀ 𝑥 ∈ 𝐴 𝐵 ⊆ 𝐶 )