Theorem rabeq12f 39069

Description: Equality deduction for restricted class abstraction. (Contributed by Giovanni Mascellani, 10-Apr-2018.)

Hypotheses

Ref	Expression
rabeq12f.1	⊢ Ⅎ𝑥𝐴
rabeq12f.2	⊢ Ⅎ𝑥𝐵

Assertion

Ref	Expression
rabeq12f	⊢ ((𝐴 = 𝐵 ∧ ∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝜓)) → {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∈ 𝐵 ∣ 𝜓})

Proof of Theorem rabeq12f

Step	Hyp	Ref	Expression
1		rabbi 3442	. . 3 ⊢ (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝜓) ↔ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∈ 𝐴 ∣ 𝜓})
2	1	biimpi 219	. 2 ⊢ (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝜓) → {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∈ 𝐴 ∣ 𝜓})
3		rabeq12f.1	. . 3 ⊢ Ⅎ𝑥𝐴
4		rabeq12f.2	. . 3 ⊢ Ⅎ𝑥𝐵
5	3, 4	rabeqf 3446	. 2 ⊢ (𝐴 = 𝐵 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐵 ∣ 𝜓})
6	2, 5	sylan9eqr 2818	1 ⊢ ((𝐴 = 𝐵 ∧ ∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝜓)) → {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∈ 𝐵 ∣ 𝜓})

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570  Ⅎwnfc 2908  ∀wral 3077  {crab 3413

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-ral 3078  df-rab 3414

This theorem is used by: (None)