Theorem equvel 2468

Description: A variable elimination law for equality with no distinct variable requirements. Compare equvini 2466. Usage of this theorem is discouraged because it depends on ax-13 2379. Use equvelv 2038 when possible. (Contributed by NM, 1-Mar-2013.) (Proof shortened by Mario Carneiro, 17-Oct-2016.) (Proof shortened by Wolf Lammen, 15-Jun-2019.) (New usage is discouraged.)

Assertion

Ref	Expression
equvel	⊢ (∀𝑧(𝑧 = 𝑥 ↔ 𝑧 = 𝑦) → 𝑥 = 𝑦)

Proof of Theorem equvel

Step	Hyp	Ref	Expression
1		albi 1820	. 2 ⊢ (∀𝑧(𝑧 = 𝑥 ↔ 𝑧 = 𝑦) → (∀𝑧 𝑧 = 𝑥 ↔ ∀𝑧 𝑧 = 𝑦))
2		ax6e 2390	. . . 4 ⊢ ∃𝑧 𝑧 = 𝑦
3		biimpr 223	. . . . . 6 ⊢ ((𝑧 = 𝑥 ↔ 𝑧 = 𝑦) → (𝑧 = 𝑦 → 𝑧 = 𝑥))
4		ax7 2023	. . . . . 6 ⊢ (𝑧 = 𝑥 → (𝑧 = 𝑦 → 𝑥 = 𝑦))
5	3, 4	syli 39	. . . . 5 ⊢ ((𝑧 = 𝑥 ↔ 𝑧 = 𝑦) → (𝑧 = 𝑦 → 𝑥 = 𝑦))
6	5	com12 32	. . . 4 ⊢ (𝑧 = 𝑦 → ((𝑧 = 𝑥 ↔ 𝑧 = 𝑦) → 𝑥 = 𝑦))
7	2, 6	eximii 1838	. . 3 ⊢ ∃𝑧((𝑧 = 𝑥 ↔ 𝑧 = 𝑦) → 𝑥 = 𝑦)
8	7	19.35i 1879	. 2 ⊢ (∀𝑧(𝑧 = 𝑥 ↔ 𝑧 = 𝑦) → ∃𝑧 𝑥 = 𝑦)
9	4	spsd 2184	. . . . 5 ⊢ (𝑧 = 𝑥 → (∀𝑧 𝑧 = 𝑦 → 𝑥 = 𝑦))
10	9	sps 2182	. . . 4 ⊢ (∀𝑧 𝑧 = 𝑥 → (∀𝑧 𝑧 = 𝑦 → 𝑥 = 𝑦))
11	10	a1dd 50	. . 3 ⊢ (∀𝑧 𝑧 = 𝑥 → (∀𝑧 𝑧 = 𝑦 → (∃𝑧 𝑥 = 𝑦 → 𝑥 = 𝑦)))
12		nfeqf 2388	. . . . 5 ⊢ ((¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦) → Ⅎ𝑧 𝑥 = 𝑦)
13	12	19.9d 2201	. . . 4 ⊢ ((¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦) → (∃𝑧 𝑥 = 𝑦 → 𝑥 = 𝑦))
14	13	ex 416	. . 3 ⊢ (¬ ∀𝑧 𝑧 = 𝑥 → (¬ ∀𝑧 𝑧 = 𝑦 → (∃𝑧 𝑥 = 𝑦 → 𝑥 = 𝑦)))
15	11, 14	bija 385	. 2 ⊢ ((∀𝑧 𝑧 = 𝑥 ↔ ∀𝑧 𝑧 = 𝑦) → (∃𝑧 𝑥 = 𝑦 → 𝑥 = 𝑦))
16	1, 8, 15	sylc 65	1 ⊢ (∀𝑧(𝑧 = 𝑥 ↔ 𝑧 = 𝑦) → 𝑥 = 𝑦)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 399  ∀wal 1536  ∃wex 1781

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-10 2142  ax-12 2175  ax-13 2379

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-nf 1786

This theorem is referenced by: (None)