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Metamath Proof Explorer

Theorem ssequn1

Description: A relationship between subclass and union. Theorem 26 of Suppes p. 27. (Contributed by NM, 30-Aug-1993) (Proof shortened by Andrew Salmon, 26-Jun-2011)

		Ref	Expression
	Assertion	ssequn1	⊢ ( 𝐴 ⊆ 𝐵 ↔ ( 𝐴 ∪ 𝐵 ) = 𝐵 )

Proof

Step	Hyp	Ref	Expression
1		bicom	⊢ ( ( 𝑥 ∈ 𝐵 ↔ ( 𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵 ) ) ↔ ( ( 𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵 ) ↔ 𝑥 ∈ 𝐵 ) )
2		pm4.72	⊢ ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐵 ↔ ( 𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵 ) ) )
3		elun	⊢ ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵 ) )
4	3	bibi1i	⊢ ( ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) ↔ 𝑥 ∈ 𝐵 ) ↔ ( ( 𝑥 ∈ 𝐴 ∨ 𝑥 ∈ 𝐵 ) ↔ 𝑥 ∈ 𝐵 ) )
5	1 2 4	3bitr4i	⊢ ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) ↔ 𝑥 ∈ 𝐵 ) )
6	5	albii	⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) ↔ ∀ 𝑥 ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) ↔ 𝑥 ∈ 𝐵 ) )
7		df-ss	⊢ ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) )
8		dfcleq	⊢ ( ( 𝐴 ∪ 𝐵 ) = 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ ( 𝐴 ∪ 𝐵 ) ↔ 𝑥 ∈ 𝐵 ) )
9	6 7 8	3bitr4i	⊢ ( 𝐴 ⊆ 𝐵 ↔ ( 𝐴 ∪ 𝐵 ) = 𝐵 )