Theorem disjif 29391

Description: Property of a disjoint collection: if 𝐵(𝑥) and 𝐵(𝑌) = 𝐷 have a common element 𝑍, then 𝑥 = 𝑌. (Contributed by Thierry Arnoux, 30-Dec-2016.)

Hypotheses

Ref	Expression
disjif.1	⊢ Ⅎ𝑥𝐶
disjif.2	⊢ (𝑥 = 𝑌 → 𝐵 = 𝐶)

Assertion

Ref	Expression
disjif	⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝑍 ∈ 𝐵 ∧ 𝑍 ∈ 𝐶)) → 𝑥 = 𝑌)

Distinct variable groups:   𝑥,𝐴   𝑥,𝑌

Allowed substitution hints:   𝐵(𝑥)   𝐶(𝑥)   𝑍(𝑥)

Proof of Theorem disjif

Step	Hyp	Ref	Expression
1		inelcm 4032	. 2 ⊢ ((𝑍 ∈ 𝐵 ∧ 𝑍 ∈ 𝐶) → (𝐵 ∩ 𝐶) ≠ ∅)
2		disjif.1	. . . . . 6 ⊢ Ⅎ𝑥𝐶
3		disjif.2	. . . . . 6 ⊢ (𝑥 = 𝑌 → 𝐵 = 𝐶)
4	2, 3	disji2f 29390	. . . . 5 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ 𝑥 ≠ 𝑌) → (𝐵 ∩ 𝐶) = ∅)
5	4	3expia 1267	. . . 4 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (𝑥 ≠ 𝑌 → (𝐵 ∩ 𝐶) = ∅))
6	5	necon1d 2816	. . 3 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → ((𝐵 ∩ 𝐶) ≠ ∅ → 𝑥 = 𝑌))
7	6	3impia 1261	. 2 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝐵 ∩ 𝐶) ≠ ∅) → 𝑥 = 𝑌)
8	1, 7	syl3an3 1361	1 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝑍 ∈ 𝐵 ∧ 𝑍 ∈ 𝐶)) → 𝑥 = 𝑌)

Syntax hints:   → wi 4   ∧ wa 384   ∧ w3a 1037   = wceq 1483   ∈ wcel 1990  Ⅎwnfc 2751   ≠ wne 2794   ∩ cin 3573  ∅c0 3915  Disj wdisj 4620

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1039  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-eu 2474  df-mo 2475  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ne 2795  df-ral 2917  df-rex 2918  df-reu 2919  df-rmo 2920  df-v 3202  df-sbc 3436  df-csb 3534  df-dif 3577  df-in 3581  df-nul 3916  df-disj 4621

This theorem is referenced by:  disjabrex  29395