Theorem equvini 2466

Description: A variable introduction law for equality. Lemma 15 of [Monk2] p. 109, however we do not require 𝑧 to be distinct from 𝑥 and 𝑦. Usage of this theorem is discouraged because it depends on ax-13 2379. See equvinv 2036 for a shorter proof requiring fewer axioms when 𝑧 is required to be distinct from 𝑥 and 𝑦. (Contributed by NM, 10-Jan-1993.) (Proof shortened by Andrew Salmon, 25-May-2011.) (Proof shortened by Wolf Lammen, 16-Sep-2023.) (New usage is discouraged.)

Assertion

Ref	Expression
equvini	⊢ (𝑥 = 𝑦 → ∃𝑧(𝑥 = 𝑧 ∧ 𝑧 = 𝑦))

Proof of Theorem equvini

Step	Hyp	Ref	Expression
1		equtr 2028	. . . 4 ⊢ (𝑧 = 𝑥 → (𝑥 = 𝑦 → 𝑧 = 𝑦))
2		equcomi 2024	. . . 4 ⊢ (𝑧 = 𝑥 → 𝑥 = 𝑧)
3	1, 2	jctild 529	. . 3 ⊢ (𝑧 = 𝑥 → (𝑥 = 𝑦 → (𝑥 = 𝑧 ∧ 𝑧 = 𝑦)))
4		19.8a 2178	. . 3 ⊢ ((𝑥 = 𝑧 ∧ 𝑧 = 𝑦) → ∃𝑧(𝑥 = 𝑧 ∧ 𝑧 = 𝑦))
5	3, 4	syl6 35	. 2 ⊢ (𝑧 = 𝑥 → (𝑥 = 𝑦 → ∃𝑧(𝑥 = 𝑧 ∧ 𝑧 = 𝑦)))
6		ax13 2382	. . 3 ⊢ (¬ 𝑧 = 𝑥 → (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
7		ax6e 2390	. . . . 5 ⊢ ∃𝑧 𝑧 = 𝑥
8	7, 3	eximii 1838	. . . 4 ⊢ ∃𝑧(𝑥 = 𝑦 → (𝑥 = 𝑧 ∧ 𝑧 = 𝑦))
9	8	19.35i 1879	. . 3 ⊢ (∀𝑧 𝑥 = 𝑦 → ∃𝑧(𝑥 = 𝑧 ∧ 𝑧 = 𝑦))
10	6, 9	syl6 35	. 2 ⊢ (¬ 𝑧 = 𝑥 → (𝑥 = 𝑦 → ∃𝑧(𝑥 = 𝑧 ∧ 𝑧 = 𝑦)))
11	5, 10	pm2.61i 185	1 ⊢ (𝑥 = 𝑦 → ∃𝑧(𝑥 = 𝑧 ∧ 𝑧 = 𝑦))

Syntax hints:  ¬ wn 3   → wi 4   ∧ 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-12 2175  ax-13 2379

This theorem depends on definitions:  df-bi 210  df-an 400  df-ex 1782

This theorem is referenced by:  2ax6elem  2482