Theorem bj-sseq 12999

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 464	. 2 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴)))
4		eqss 3112	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5	3, 4	syl6bbr 197	1 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) ↔ 𝐴 = 𝐵))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104   = wceq 1331   ⊆ wss 3071

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-11 1484  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121

This theorem depends on definitions:  df-bi 116  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-in 3077  df-ss 3084

This theorem is referenced by: (None)