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Theorem s3eqd 14913
Description: Equality theorem for a length 3 word. (Contributed by Mario Carneiro, 27-Feb-2016.)
Hypotheses
Ref Expression
s2eqd.1 (𝜑𝐴 = 𝑁)
s2eqd.2 (𝜑𝐵 = 𝑂)
s3eqd.3 (𝜑𝐶 = 𝑃)
Assertion
Ref Expression
s3eqd (𝜑 → ⟨“𝐴𝐵𝐶”⟩ = ⟨“𝑁𝑂𝑃”⟩)

Proof of Theorem s3eqd
StepHypRef Expression
1 s2eqd.1 . . . 4 (𝜑𝐴 = 𝑁)
2 s2eqd.2 . . . 4 (𝜑𝐵 = 𝑂)
31, 2s2eqd 14912 . . 3 (𝜑 → ⟨“𝐴𝐵”⟩ = ⟨“𝑁𝑂”⟩)
4 s3eqd.3 . . . 4 (𝜑𝐶 = 𝑃)
54s1eqd 14649 . . 3 (𝜑 → ⟨“𝐶”⟩ = ⟨“𝑃”⟩)
63, 5oveq12d 7466 . 2 (𝜑 → (⟨“𝐴𝐵”⟩ ++ ⟨“𝐶”⟩) = (⟨“𝑁𝑂”⟩ ++ ⟨“𝑃”⟩))
7 df-s3 14898 . 2 ⟨“𝐴𝐵𝐶”⟩ = (⟨“𝐴𝐵”⟩ ++ ⟨“𝐶”⟩)
8 df-s3 14898 . 2 ⟨“𝑁𝑂𝑃”⟩ = (⟨“𝑁𝑂”⟩ ++ ⟨“𝑃”⟩)
96, 7, 83eqtr4g 2805 1 (𝜑 → ⟨“𝐴𝐵𝐶”⟩ = ⟨“𝑁𝑂𝑃”⟩)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1537  (class class class)co 7448   ++ cconcat 14618  ⟨“cs1 14643  ⟨“cs2 14890  ⟨“cs3 14891
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-3an 1089  df-tru 1540  df-fal 1550  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-rab 3444  df-v 3490  df-dif 3979  df-un 3981  df-ss 3993  df-nul 4353  df-if 4549  df-sn 4649  df-pr 4651  df-op 4655  df-uni 4932  df-br 5167  df-iota 6525  df-fv 6581  df-ov 7451  df-s1 14644  df-s2 14897  df-s3 14898
This theorem is referenced by:  s4eqd  14914  s3eq2  14919  s3sndisj  15016  s3iunsndisj  15017  ragcgr  28733  perpneq  28740  isperp2  28741  isperp2d  28742  footexALT  28744  footexlem2  28746  foot  28748  perprag  28752  perpdragALT  28753  colperpexlem1  28756  lmiisolem  28822  hypcgrlem1  28825  hypcgrlem2  28826  trgcopyeu  28832  iscgra  28835  iscgra1  28836  iscgrad  28837  sacgr  28857  isleag  28873  isleagd  28874  iseqlg  28893
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