Theorem epinid0 9603

Description: The membership relation and the identity relation are disjoint. Variable-free version of nelaneq 9600. (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 5550	. . 3 ⊢ E = {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦}
2		df-id 5545	. . 3 ⊢ I = {⟨𝑥, 𝑦⟩ ∣ 𝑥 = 𝑦}
3	1, 2	ineq12i 4115	. 2 ⊢ ( E ∩ I ) = ({⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦} ∩ {⟨𝑥, 𝑦⟩ ∣ 𝑥 = 𝑦})
4		inopab 5806	. 2 ⊢ ({⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦} ∩ {⟨𝑥, 𝑦⟩ ∣ 𝑥 = 𝑦}) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝑦 ∧ 𝑥 = 𝑦)}
5		nelaneq 9600	. . . 4 ⊢ ¬ (𝑥 ∈ 𝑦 ∧ 𝑥 = 𝑦)
6	5	gen2 1828	. . 3 ⊢ ∀𝑥∀𝑦 ¬ (𝑥 ∈ 𝑦 ∧ 𝑥 = 𝑦)
7		opab0 5528	. . 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 2787	1 ⊢ ( E ∩ I ) = ∅

This proof depends on syntax axioms:  ¬ wn 3   ∧ wa 401  ∀wal 1567   = wceq 1569   ∩ cin 4089  ∅c0 4278  {copab 5166   I cid 5544   E cep 5549

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 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732  ax-sep 5248  ax-pr 5390  ax-reg 9590

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ne 2956  df-rab 3413  df-v 3452  df-ss 3896  df-dif 4047  df-in 4091  df-un 4151  df-nul 4279  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-opab 5167  df-id 5545  df-eprel 5550  df-xp 5656  df-rel 5657

This theorem is used by: (None)