Theorem bj-gabss 34809

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 2740	. . . . . . . 8 ⊢ (𝐴 = 𝐵 → (𝐴 = 𝑦 ↔ 𝐵 = 𝑦))
2	1	biimpd 232	. . . . . . 7 ⊢ (𝐴 = 𝐵 → (𝐴 = 𝑦 → 𝐵 = 𝑦))
3	2	adantr 484	. . . . . 6 ⊢ ((𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → (𝐴 = 𝑦 → 𝐵 = 𝑦))
4		simpr 488	. . . . . 6 ⊢ ((𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → (𝜑 → 𝜓))
5	3, 4	anim12d 612	. . . . 5 ⊢ ((𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → ((𝐴 = 𝑦 ∧ 𝜑) → (𝐵 = 𝑦 ∧ 𝜓)))
6	5	aleximi 1839	. . . 4 ⊢ (∀𝑥(𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → (∃𝑥(𝐴 = 𝑦 ∧ 𝜑) → ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)))
7	6	alrimiv 1935	. . 3 ⊢ (∀𝑥(𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → ∀𝑦(∃𝑥(𝐴 = 𝑦 ∧ 𝜑) → ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)))
8		ss2ab 3959	. . 3 ⊢ ({𝑦 ∣ ∃𝑥(𝐴 = 𝑦 ∧ 𝜑)} ⊆ {𝑦 ∣ ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)} ↔ ∀𝑦(∃𝑥(𝐴 = 𝑦 ∧ 𝜑) → ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)))
9	7, 8	sylibr 237	. 2 ⊢ (∀𝑥(𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → {𝑦 ∣ ∃𝑥(𝐴 = 𝑦 ∧ 𝜑)} ⊆ {𝑦 ∣ ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)})
10		df-bj-gab 34808	. 2 ⊢ {𝐴 ∣ 𝑥 ∣ 𝜑} = {𝑦 ∣ ∃𝑥(𝐴 = 𝑦 ∧ 𝜑)}
11		df-bj-gab 34808	. 2 ⊢ {𝐵 ∣ 𝑥 ∣ 𝜓} = {𝑦 ∣ ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)}
12	9, 10, 11	3sstr4g 3932	1 ⊢ (∀𝑥(𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → {𝐴 ∣ 𝑥 ∣ 𝜑} ⊆ {𝐵 ∣ 𝑥 ∣ 𝜓})

Syntax hints:   → wi 4   ∧ wa 399  ∀wal 1541   = wceq 1543  ∃wex 1787  {cab 2714   ⊆ wss 3853  {bj-cgab 34807

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2018  ax-8 2114  ax-9 2122  ax-10 2143  ax-11 2160  ax-12 2177  ax-ext 2708

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 848  df-tru 1546  df-ex 1788  df-nf 1792  df-sb 2073  df-clab 2715  df-cleq 2728  df-clel 2809  df-nfc 2879  df-v 3400  df-in 3860  df-ss 3870  df-bj-gab 34808

This theorem is referenced by:  bj-gabssd  34810