Theorem disjif 32558

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 4412	. 2 ⊢ ((𝑍 ∈ 𝐵 ∧ 𝑍 ∈ 𝐶) → (𝐵 ∩ 𝐶) ≠ ∅)
2		disjif.1	. . . . . 6 ⊢ Ⅎ𝑥𝐶
3		disjif.2	. . . . . 6 ⊢ (𝑥 = 𝑌 → 𝐵 = 𝐶)
4	2, 3	disji2f 32557	. . . . 5 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ 𝑥 ≠ 𝑌) → (𝐵 ∩ 𝐶) = ∅)
5	4	3expia 1121	. . . 4 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (𝑥 ≠ 𝑌 → (𝐵 ∩ 𝐶) = ∅))
6	5	necon1d 2950	. . 3 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → ((𝐵 ∩ 𝐶) ≠ ∅ → 𝑥 = 𝑌))
7	6	3impia 1117	. 2 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝐵 ∩ 𝐶) ≠ ∅) → 𝑥 = 𝑌)
8	1, 7	syl3an3 1165	1 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝑍 ∈ 𝐵 ∧ 𝑍 ∈ 𝐶)) → 𝑥 = 𝑌)

Syntax hints:   → wi 4   ∧ wa 395   ∧ w3a 1086   = wceq 1541   ∈ wcel 2111  Ⅎwnfc 2879   ≠ wne 2928   ∩ cin 3896  ∅c0 4280  Disj wdisj 5056

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-10 2144  ax-11 2160  ax-12 2180  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-nf 1785  df-sb 2068  df-mo 2535  df-clab 2710  df-cleq 2723  df-clel 2806  df-nfc 2881  df-ne 2929  df-ral 3048  df-rmo 3346  df-rab 3396  df-v 3438  df-sbc 3737  df-csb 3846  df-dif 3900  df-in 3904  df-nul 4281  df-disj 5057

This theorem is referenced by:  disjabrex  32562  2ndresdju  32631