Theorem epinid0 9577

Description: The membership relation and the identity relation are disjoint. Variable-free version of nelaneq 9574. (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 5566	. . 3 ⊢ E = {⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦}
2		df-id 5561	. . 3 ⊢ I = {⟨𝑥, 𝑦⟩ ∣ 𝑥 = 𝑦}
3	1, 2	ineq12i 4174	. 2 ⊢ ( E ∩ I ) = ({⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦} ∩ {⟨𝑥, 𝑦⟩ ∣ 𝑥 = 𝑦})
4		inopab 5821	. 2 ⊢ ({⟨𝑥, 𝑦⟩ ∣ 𝑥 ∈ 𝑦} ∩ {⟨𝑥, 𝑦⟩ ∣ 𝑥 = 𝑦}) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝑦 ∧ 𝑥 = 𝑦)}
5		nelaneq 9574	. . . 4 ⊢ ¬ (𝑥 ∈ 𝑦 ∧ 𝑥 = 𝑦)
6	5	gen2 1829	. . 3 ⊢ ∀𝑥∀𝑦 ¬ (𝑥 ∈ 𝑦 ∧ 𝑥 = 𝑦)
7		opab0 5544	. . 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 2793	1 ⊢ ( E ∩ I ) = ∅

This proof depends on syntax axioms:  ¬ wn 3   ∧ wa 401  ∀wal 1568   = wceq 1570   ∩ cin 3907  ∅c0 4289  {copab 5178   I cid 5560   E cep 5565

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 2148  ax-9 2156  ax-10 2179  ax-11 2195  ax-12 2216  ax-ext 2738  ax-sep 5262  ax-pr 5409  ax-reg 9564

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 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-ne 2962  df-rab 3420  df-v 3460  df-dif 3911  df-un 3913  df-in 3915  df-ss 3925  df-nul 4290  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-opab 5179  df-id 5561  df-eprel 5566  df-xp 5672  df-rel 5673

This theorem is used by: (None)