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Theorem equvel 2459
Description: A variable elimination law for equality with no distinct variable requirements. Compare equvini 2458. Usage of this theorem is discouraged because it depends on ax-13 2375. Use equvelv 2028 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 = 𝑥𝑧 = 𝑦) → 𝑥 = 𝑦)

Proof of Theorem equvel
StepHypRef Expression
1 albi 1815 . 2 (∀𝑧(𝑧 = 𝑥𝑧 = 𝑦) → (∀𝑧 𝑧 = 𝑥 ↔ ∀𝑧 𝑧 = 𝑦))
2 ax6e 2386 . . . 4 𝑧 𝑧 = 𝑦
3 biimpr 220 . . . . . 6 ((𝑧 = 𝑥𝑧 = 𝑦) → (𝑧 = 𝑦𝑧 = 𝑥))
4 ax7 2013 . . . . . 6 (𝑧 = 𝑥 → (𝑧 = 𝑦𝑥 = 𝑦))
53, 4syli 39 . . . . 5 ((𝑧 = 𝑥𝑧 = 𝑦) → (𝑧 = 𝑦𝑥 = 𝑦))
65com12 32 . . . 4 (𝑧 = 𝑦 → ((𝑧 = 𝑥𝑧 = 𝑦) → 𝑥 = 𝑦))
72, 6eximii 1834 . . 3 𝑧((𝑧 = 𝑥𝑧 = 𝑦) → 𝑥 = 𝑦)
8719.35i 1876 . 2 (∀𝑧(𝑧 = 𝑥𝑧 = 𝑦) → ∃𝑧 𝑥 = 𝑦)
94spsd 2185 . . . . 5 (𝑧 = 𝑥 → (∀𝑧 𝑧 = 𝑦𝑥 = 𝑦))
109sps 2183 . . . 4 (∀𝑧 𝑧 = 𝑥 → (∀𝑧 𝑧 = 𝑦𝑥 = 𝑦))
1110a1dd 50 . . 3 (∀𝑧 𝑧 = 𝑥 → (∀𝑧 𝑧 = 𝑦 → (∃𝑧 𝑥 = 𝑦𝑥 = 𝑦)))
12 nfeqf 2384 . . . . 5 ((¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦) → Ⅎ𝑧 𝑥 = 𝑦)
131219.9d 2201 . . . 4 ((¬ ∀𝑧 𝑧 = 𝑥 ∧ ¬ ∀𝑧 𝑧 = 𝑦) → (∃𝑧 𝑥 = 𝑦𝑥 = 𝑦))
1413ex 412 . . 3 (¬ ∀𝑧 𝑧 = 𝑥 → (¬ ∀𝑧 𝑧 = 𝑦 → (∃𝑧 𝑥 = 𝑦𝑥 = 𝑦)))
1511, 14bija 380 . 2 ((∀𝑧 𝑧 = 𝑥 ↔ ∀𝑧 𝑧 = 𝑦) → (∃𝑧 𝑥 = 𝑦𝑥 = 𝑦))
161, 8, 15sylc 65 1 (∀𝑧(𝑧 = 𝑥𝑧 = 𝑦) → 𝑥 = 𝑦)
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 206  wa 395  wal 1535  wex 1776
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-10 2139  ax-12 2175  ax-13 2375
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1540  df-ex 1777  df-nf 1781
This theorem is referenced by: (None)
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