Theorem bj-gabss 37599

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 2767	. . . . . . . 8 ⊢ (𝐴 = 𝐵 → (𝐴 = 𝑦 ↔ 𝐵 = 𝑦))
2	1	biimpd 232	. . . . . . 7 ⊢ (𝐴 = 𝐵 → (𝐴 = 𝑦 → 𝐵 = 𝑦))
3	2	adantr 485	. . . . . 6 ⊢ ((𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → (𝐴 = 𝑦 → 𝐵 = 𝑦))
4		simpr 489	. . . . . 6 ⊢ ((𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → (𝜑 → 𝜓))
5	3, 4	anim12d 620	. . . . 5 ⊢ ((𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → ((𝐴 = 𝑦 ∧ 𝜑) → (𝐵 = 𝑦 ∧ 𝜓)))
6	5	aleximi 1862	. . . 4 ⊢ (∀𝑥(𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → (∃𝑥(𝐴 = 𝑦 ∧ 𝜑) → ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)))
7	6	alrimiv 1957	. . 3 ⊢ (∀𝑥(𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → ∀𝑦(∃𝑥(𝐴 = 𝑦 ∧ 𝜑) → ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)))
8		ss2ab 4015	. . 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 37598	. 2 ⊢ {𝐴 ∣ 𝑥 ∣ 𝜑} = {𝑦 ∣ ∃𝑥(𝐴 = 𝑦 ∧ 𝜑)}
11		df-bj-gab 37598	. 2 ⊢ {𝐵 ∣ 𝑥 ∣ 𝜓} = {𝑦 ∣ ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)}
12	9, 10, 11	3sstr4g 3990	1 ⊢ (∀𝑥(𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → {𝐴 ∣ 𝑥 ∣ 𝜑} ⊆ {𝐵 ∣ 𝑥 ∣ 𝜓})

This proof depends on syntax axioms:   → wi 4   ∧ wa 400  ∀wal 1568   = wceq 1570  ∃wex 1809  {cab 2741   ⊆ wss 3905  {bj-cgab 37597

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-10 2176  ax-11 2192  ax-12 2213  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-ex 1810  df-nf 1814  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-nfc 2912  df-ss 3922  df-bj-gab 37598

This theorem is used by:  bj-gabssd  37600