Theorem epinid0 9629

Description: The membership relation and the identity relation are disjoint. Variable-free version of nelaneq 9626. (Proposed by BJ, 18-Jun-2022.) (Contributed by AV, 18-Jun-2022.)

Assertion

Ref	Expression
epinid0	⊢ ( E ∩ I ) = ∅

Proof of Theorem epinid0

Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-eprel 5556	. . 3 ⊢ E = {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦}
2		df-id 5551	. . 3 ⊢ I = {⟨𝑥, 𝑦⟩ ∣ 𝑥 = 𝑦}
3	1, 2	ineq12i 4118	. 2 ⊢ ( E ∩ I ) = ({⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦} ∩ {⟨𝑥, 𝑦⟩ ∣ 𝑥 = 𝑦})
4		inopab 5814	. 2 ⊢ ({⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦} ∩ {⟨𝑥, 𝑦⟩ ∣ 𝑥 = 𝑦}) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝑦 ∧ 𝑥 = 𝑦)}
5		nelaneq 9626	. . . 4 ⊢ ¬ (𝑥 ∈ 𝑦 ∧ 𝑥 = 𝑦)
6	5	gen2 1829	. . 3 ⊢ ∀𝑥∀𝑦 ¬ (𝑥 ∈ 𝑦 ∧ 𝑥 = 𝑦)
7		opab0 5534	. . 3 ⊢ ({⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝑦 ∧ 𝑥 = 𝑦)} = ∅ ↔ ∀𝑥∀𝑦 ¬ (𝑥 ∈ 𝑦 ∧ 𝑥 = 𝑦))
8	6, 7	mpbir 234	. 2 ⊢ {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝑦 ∧ 𝑥 = 𝑦)} = ∅
9	3, 4, 8	3eqtri 2790	1 ⊢ ( E ∩ I ) = ∅

This proof depends on syntax axioms:  ¬ wn 3   ∧ wa 401  ∀wal 1568   = wceq 1570   ∩ cin 4092  ∅c0 4281  {copab 5171   I cid 5550   E cep 5555

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735  ax-sep 5253  ax-pr 5396  ax-reg 9616

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ne 2960  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-opab 5172  df-id 5551  df-eprel 5556  df-xp 5662  df-rel 5663

This theorem is used by: (None)