Theorem rabeqel 39109

Description: Class element of a restricted class abstraction. (Contributed by Peter Mazsa, 24-Jul-2021.)

Hypotheses

Ref	Expression
rabeqel.1	⊢ 𝐵 = {𝑥 ∈ 𝐴 ∣ 𝜑}
rabeqel.2	⊢ (𝑥 = 𝐶 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
rabeqel	⊢ (𝐶 ∈ 𝐵 ↔ (𝜓 ∧ 𝐶 ∈ 𝐴))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐶   𝜓,𝑥

Allowed substitution hints:   𝜑(𝑥)   𝐵(𝑥)

Proof of Theorem rabeqel

Step	Hyp	Ref	Expression
1		rabeqel.2	. . 3 ⊢ (𝑥 = 𝐶 → (𝜑 ↔ 𝜓))
2		rabeqel.1	. . 3 ⊢ 𝐵 = {𝑥 ∈ 𝐴 ∣ 𝜑}
3	1, 2	elrab2 3648	. 2 ⊢ (𝐶 ∈ 𝐵 ↔ (𝐶 ∈ 𝐴 ∧ 𝜓))
4	3	biancomi 468	1 ⊢ (𝐶 ∈ 𝐵 ↔ (𝜓 ∧ 𝐶 ∈ 𝐴))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2145  {crab 3412

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-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452

This theorem is used by:  elrefrels2  39450  elrefrels3  39451  elcnvrefrels2  39466  elcnvrefrels3  39467  elsymrels2  39489  elsymrels3  39490  elsymrels4  39491  elsymrels5  39492  elrefsymrels2  39505  eltrrels2  39515  eltrrels3  39516  eleqvrels2  39528  eleqvrels3  39529  elfunsALTV  39629  eldisjs  39671