Theorem bj-inrab2 37821

Description: Shorter proof of inrab 4262. (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 37820	. 2 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ (𝐴 ∩ 𝐴) ∣ (𝜑 ∧ 𝜓)}
2		nfv 1947	. . . 4 ⊢ Ⅎ𝑥⊤
3		inidm 4172	. . . . 5 ⊢ (𝐴 ∩ 𝐴) = 𝐴
4	3	a1i 11	. . . 4 ⊢ (⊤ → (𝐴 ∩ 𝐴) = 𝐴)
5	2, 4	rabeqd 3440	. . 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 2784	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∩ {𝑥 ∈ 𝐴 ∣ 𝜓}) = {𝑥 ∈ 𝐴 ∣ (𝜑 ∧ 𝜓)}

This proof depends on syntax axioms:   ∧ wa 401   = wceq 1570  ⊤wtru 1571  {crab 3413   ∩ cin 3898

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-12 2213  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-rab 3414  df-v 3453  df-in 3906

This theorem is used by: (None)