Theorem bj-inrab2 35808

Description: Shorter proof of inrab 4307. (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 35807	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ (𝐴 ∩ 𝐴) ∣ (𝜑 ∧ 𝜓)}
2		nfv 1918	. . . 4 ⊢ Ⅎ𝑥⊤
3		inidm 4219	. . . . 5 ⊢ (𝐴 ∩ 𝐴) = 𝐴
4	3	a1i 11	. . . 4 ⊢ (⊤ → (𝐴 ∩ 𝐴) = 𝐴)
5	2, 4	rabeqd 3461	. . 3 ⊢ (⊤ → {𝑥 ∈ (𝐴 ∩ 𝐴) ∣ (𝜑 ∧ 𝜓)} = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)})
6	5	mptru 1549	. 2 ⊢ {𝑥 ∈ (𝐴 ∩ 𝐴) ∣ (𝜑 ∧ 𝜓)} = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}
7	1, 6	eqtri 2761	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}

Syntax hints:   ∧ wa 397   = wceq 1542  ⊤wtru 1543  {crab 3433   ∩ cin 3948

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-12 2172  ax-ext 2704

This theorem depends on definitions:  df-bi 206  df-an 398  df-tru 1545  df-ex 1783  df-nf 1787  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-rab 3434  df-v 3477  df-in 3956

This theorem is referenced by: (None)