Theorem bj-inrab2 37564

Description: Shorter proof of inrab 4269. (Contributed by BJ, 21-Apr-2019.) (Proof modification is discouraged.)

Assertion

Ref	Expression
bj-inrab2	⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}

Proof of Theorem bj-inrab2

Step	Hyp	Ref	Expression
1		bj-inrab 37563	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ (𝐴 ∩ 𝐴) ∣ (𝜑 ∧ 𝜓)}
2		nfv 1944	. . . 4 ⊢ Ⅎ𝑥⊤
3		inidm 4179	. . . . 5 ⊢ (𝐴 ∩ 𝐴) = 𝐴
4	3	a1i 11	. . . 4 ⊢ (⊤ → (𝐴 ∩ 𝐴) = 𝐴)
5	2, 4	rabeqd 3444	. . 3 ⊢ (⊤ → {𝑥 ∈ (𝐴 ∩ 𝐴) ∣ (𝜑 ∧ 𝜓)} = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)})
6	5	mptru 1577	. 2 ⊢ {𝑥 ∈ (𝐴 ∩ 𝐴) ∣ (𝜑 ∧ 𝜓)} = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}
7	1, 6	eqtri 2786	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}

Syntax hints:   ∧ wa 400   = wceq 1570  ⊤wtru 1571  {crab 3416   ∩ cin 3904

This theorem was proved from 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-12 2213  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1573  df-ex 1810  df-nf 1814  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-rab 3417  df-v 3457  df-in 3912

This theorem is referenced by: (None)