Theorem bj-gabss 36118

Description: Inclusion of generalized class abstractions. (Contributed by BJ, 4-Oct-2024.)

Assertion

Ref	Expression
bj-gabss	⊢ (∀𝑥(𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → {𝐴 ∣ 𝑥 ∣ 𝜑} ⊆ {𝐵 ∣ 𝑥 ∣ 𝜓})

Proof of Theorem bj-gabss

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eqeq1 2734	. . . . . . . 8 ⊢ (𝐴 = 𝐵 → (𝐴 = 𝑦 ↔ 𝐵 = 𝑦))
2	1	biimpd 228	. . . . . . 7 ⊢ (𝐴 = 𝐵 → (𝐴 = 𝑦 → 𝐵 = 𝑦))
3	2	adantr 479	. . . . . 6 ⊢ ((𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → (𝐴 = 𝑦 → 𝐵 = 𝑦))
4		simpr 483	. . . . . 6 ⊢ ((𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → (𝜑 → 𝜓))
5	3, 4	anim12d 607	. . . . 5 ⊢ ((𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → ((𝐴 = 𝑦 ∧ 𝜑) → (𝐵 = 𝑦 ∧ 𝜓)))
6	5	aleximi 1832	. . . 4 ⊢ (∀𝑥(𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → (∃𝑥(𝐴 = 𝑦 ∧ 𝜑) → ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)))
7	6	alrimiv 1928	. . 3 ⊢ (∀𝑥(𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → ∀𝑦(∃𝑥(𝐴 = 𝑦 ∧ 𝜑) → ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)))
8		ss2ab 4055	. . 3 ⊢ ({𝑦 ∣ ∃𝑥(𝐴 = 𝑦 ∧ 𝜑)} ⊆ {𝑦 ∣ ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)} ↔ ∀𝑦(∃𝑥(𝐴 = 𝑦 ∧ 𝜑) → ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)))
9	7, 8	sylibr 233	. 2 ⊢ (∀𝑥(𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → {𝑦 ∣ ∃𝑥(𝐴 = 𝑦 ∧ 𝜑)} ⊆ {𝑦 ∣ ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)})
10		df-bj-gab 36117	. 2 ⊢ {𝐴 ∣ 𝑥 ∣ 𝜑} = {𝑦 ∣ ∃𝑥(𝐴 = 𝑦 ∧ 𝜑)}
11		df-bj-gab 36117	. 2 ⊢ {𝐵 ∣ 𝑥 ∣ 𝜓} = {𝑦 ∣ ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)}
12	9, 10, 11	3sstr4g 4026	1 ⊢ (∀𝑥(𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → {𝐴 ∣ 𝑥 ∣ 𝜑} ⊆ {𝐵 ∣ 𝑥 ∣ 𝜓})

Syntax hints:   → wi 4   ∧ wa 394  ∀wal 1537   = wceq 1539  ∃wex 1779  {cab 2707   ⊆ wss 3947  {bj-cgab 36116

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-10 2135  ax-11 2152  ax-12 2169  ax-ext 2701

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 844  df-tru 1542  df-ex 1780  df-nf 1784  df-sb 2066  df-clab 2708  df-cleq 2722  df-clel 2808  df-nfc 2883  df-v 3474  df-in 3954  df-ss 3964  df-bj-gab 36117

This theorem is referenced by:  bj-gabssd  36119