Theorem bj-sseq 15862

Description: If two converse inclusions are characterized each by a formula, then equality is characterized by the conjunction of these formulas. (Contributed by BJ, 30-Nov-2019.)

Hypotheses

Ref	Expression
bj-sseq.1	⊢ (𝜑 → (𝜓 ↔ 𝐴 ⊆ 𝐵))
bj-sseq.2	⊢ (𝜑 → (𝜒 ↔ 𝐵 ⊆ 𝐴))

Assertion

Ref	Expression
bj-sseq	⊢ (𝜑 → ((𝜓 ∧ 𝜒) ↔ 𝐴 = 𝐵))

Proof of Theorem bj-sseq

Step	Hyp	Ref	Expression
1		bj-sseq.1	. . 3 ⊢ (𝜑 → (𝜓 ↔ 𝐴 ⊆ 𝐵))
2		bj-sseq.2	. . 3 ⊢ (𝜑 → (𝜒 ↔ 𝐵 ⊆ 𝐴))
3	1, 2	anbi12d 473	. 2 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴)))
4		eqss 3212	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5	3, 4	bitr4di 198	1 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) ↔ 𝐴 = 𝐵))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1373   ⊆ wss 3170

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-11 1530  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-in 3176  df-ss 3183

This theorem is referenced by: (None)