Theorem disjif 32498

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 4459	. 2 ⊢ ((𝑍 ∈ 𝐵 ∧ 𝑍 ∈ 𝐶) → (𝐵 ∩ 𝐶) ≠ ∅)
2		disjif.1	. . . . . 6 ⊢ Ⅎ𝑥𝐶
3		disjif.2	. . . . . 6 ⊢ (𝑥 = 𝑌 → 𝐵 = 𝐶)
4	2, 3	disji2f 32497	. . . . 5 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ 𝑥 ≠ 𝑌) → (𝐵 ∩ 𝐶) = ∅)
5	4	3expia 1118	. . . 4 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (𝑥 ≠ 𝑌 → (𝐵 ∩ 𝐶) = ∅))
6	5	necon1d 2952	. . 3 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → ((𝐵 ∩ 𝐶) ≠ ∅ → 𝑥 = 𝑌))
7	6	3impia 1114	. 2 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝐵 ∩ 𝐶) ≠ ∅) → 𝑥 = 𝑌)
8	1, 7	syl3an3 1162	1 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝑍 ∈ 𝐵 ∧ 𝑍 ∈ 𝐶)) → 𝑥 = 𝑌)

Syntax hints:   → wi 4   ∧ wa 394   ∧ w3a 1084   = wceq 1534   ∈ wcel 2099  Ⅎwnfc 2876   ≠ wne 2930   ∩ cin 3945  ∅c0 4322  Disj wdisj 5110

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-10 2130  ax-11 2147  ax-12 2167  ax-ext 2697

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-3an 1086  df-tru 1537  df-fal 1547  df-ex 1775  df-nf 1779  df-sb 2061  df-mo 2529  df-clab 2704  df-cleq 2718  df-clel 2803  df-nfc 2878  df-ne 2931  df-ral 3052  df-rmo 3364  df-rab 3420  df-v 3464  df-sbc 3776  df-csb 3892  df-dif 3949  df-in 3953  df-nul 4323  df-disj 5111

This theorem is referenced by:  disjabrex  32502  2ndresdju  32566