Theorem bj-sseq 9931

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 442	. 2 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴)))
4		eqss 2960	. 2 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5	3, 4	syl6bbr 187	1 ⊢ (𝜑 → ((𝜓 ∧ 𝜒) ↔ 𝐴 = 𝐵))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98   = wceq 1243   ⊆ wss 2917

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-11 1397  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-in 2924  df-ss 2931

This theorem is referenced by: (None)