Theorem equvini 2484

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 2401. See equvinv 2061 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 2053	. . . 4 ⊢ (𝑧 = 𝑥 → (𝑥 = 𝑦 → 𝑧 = 𝑦))
2		equcomi 2049	. . . 4 ⊢ (𝑧 = 𝑥 → 𝑥 = 𝑧)
3	1, 2	jctild 535	. . 3 ⊢ (𝑧 = 𝑥 → (𝑥 = 𝑦 → (𝑥 = 𝑧 ∧ 𝑧 = 𝑦)))
4		19.8a 2217	. . 3 ⊢ ((𝑥 = 𝑧 ∧ 𝑧 = 𝑦) → ∃𝑧(𝑥 = 𝑧 ∧ 𝑧 = 𝑦))
5	3, 4	syl6 36	. 2 ⊢ (𝑧 = 𝑥 → (𝑥 = 𝑦 → ∃𝑧(𝑥 = 𝑧 ∧ 𝑧 = 𝑦)))
6		ax13 2404	. . 3 ⊢ (¬ 𝑧 = 𝑥 → (𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
7		ax6e 2412	. . . . 5 ⊢ ∃𝑧 𝑧 = 𝑥
8	7, 3	eximii 1869	. . . 4 ⊢ ∃𝑧(𝑥 = 𝑦 → (𝑥 = 𝑧 ∧ 𝑧 = 𝑦))
9	8	19.35i 1910	. . 3 ⊢ (∀𝑧 𝑥 = 𝑦 → ∃𝑧(𝑥 = 𝑧 ∧ 𝑧 = 𝑦))
10	6, 9	syl6 36	. 2 ⊢ (¬ 𝑧 = 𝑥 → (𝑥 = 𝑦 → ∃𝑧(𝑥 = 𝑧 ∧ 𝑧 = 𝑦)))
11	5, 10	pm2.61i 184	1 ⊢ (𝑥 = 𝑦 → ∃𝑧(𝑥 = 𝑧 ∧ 𝑧 = 𝑦))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401  ∀wal 1567  ∃wex 1811

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-12 2212  ax-13 2401

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812

This theorem is used by:  2ax6elem  2499