Theorem ineqri 4112

Description: Inference from membership to intersection. (Contributed by NM, 21-Jun-1993.)

Hypothesis

Ref	Expression
ineqri.1	⊢ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ 𝐶)

Assertion

Ref	Expression
ineqri	⊢ (𝐴 ∩ 𝐵) = 𝐶

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

Proof of Theorem ineqri

Step	Hyp	Ref	Expression
1		elin 4096	. . 3 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
2		ineqri.1	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ 𝐶)
3	1, 2	bitri 278	. 2 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ 𝑥 ∈ 𝐶)
4	3	eqriv 2760	1 ⊢ (𝐴 ∩ 𝐵) = 𝐶

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1569   ∈ wcel 2146   ∩ cin 4092

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456  df-in 4094

This theorem is used by:  inidm  4126  inass  4127  dfin2  4219  indi  4232  inab  4257  in0  4348  pwin  5546  dfres3  5980  dmres  6009  inixp  38808