Theorem bj-inrab2 37410

Description: Shorter proof of inrab 4268. (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 37409	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ (𝐴 ∩ 𝐴) ∣ (𝜑 ∧ 𝜓)}
2		nfv 1934	. . . 4 ⊢ Ⅎ𝑥⊤
3		inidm 4178	. . . . 5 ⊢ (𝐴 ∩ 𝐴) = 𝐴
4	3	a1i 11	. . . 4 ⊢ (⊤ → (𝐴 ∩ 𝐴) = 𝐴)
5	2, 4	rabeqd 3442	. . 3 ⊢ (⊤ → {𝑥 ∈ (𝐴 ∩ 𝐴) ∣ (𝜑 ∧ 𝜓)} = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)})
6	5	mptru 1567	. 2 ⊢ {𝑥 ∈ (𝐴 ∩ 𝐴) ∣ (𝜑 ∧ 𝜓)} = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}
7	1, 6	eqtri 2785	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}

Syntax hints:   ∧ wa 399   = wceq 1560  ⊤wtru 1561  {crab 3414   ∩ cin 3903

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-12 2212  ax-ext 2734

This theorem depends on definitions:  df-bi 209  df-an 400  df-tru 1563  df-ex 1800  df-nf 1804  df-sb 2091  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3415  df-v 3456  df-in 3911

This theorem is referenced by: (None)