Theorem s3eqd 14900

Description: Equality theorem for a length 3 word. (Contributed by Mario Carneiro, 27-Feb-2016.)

Hypotheses

Ref	Expression
s2eqd.1	⊢ (𝜑 → 𝐴 = 𝑁)
s2eqd.2	⊢ (𝜑 → 𝐵 = 𝑂)
s3eqd.3	⊢ (𝜑 → 𝐶 = 𝑃)

Assertion

Ref	Expression
s3eqd	⊢ (𝜑 → ⟨“𝐴𝐵𝐶”⟩ = ⟨“𝑁𝑂𝑃”⟩)

Proof of Theorem s3eqd

Step	Hyp	Ref	Expression
1		s2eqd.1	. . . 4 ⊢ (𝜑 → 𝐴 = 𝑁)
2		s2eqd.2	. . . 4 ⊢ (𝜑 → 𝐵 = 𝑂)
3	1, 2	s2eqd 14899	. . 3 ⊢ (𝜑 → ⟨“𝐴𝐵”⟩ = ⟨“𝑁𝑂”⟩)
4		s3eqd.3	. . . 4 ⊢ (𝜑 → 𝐶 = 𝑃)
5	4	s1eqd 14636	. . 3 ⊢ (𝜑 → ⟨“𝐶”⟩ = ⟨“𝑃”⟩)
6	3, 5	oveq12d 7449	. 2 ⊢ (𝜑 → (⟨“𝐴𝐵”⟩ ++ ⟨“𝐶”⟩) = (⟨“𝑁𝑂”⟩ ++ ⟨“𝑃”⟩))
7		df-s3 14885	. 2 ⊢ ⟨“𝐴𝐵𝐶”⟩ = (⟨“𝐴𝐵”⟩ ++ ⟨“𝐶”⟩)
8		df-s3 14885	. 2 ⊢ ⟨“𝑁𝑂𝑃”⟩ = (⟨“𝑁𝑂”⟩ ++ ⟨“𝑃”⟩)
9	6, 7, 8	3eqtr4g 2800	1 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶”⟩ = ⟨“𝑁𝑂𝑃”⟩)

Syntax hints:   → wi 4   = wceq 1537  (class class class)co 7431   ++ cconcat 14605  ⟨“cs1 14630  ⟨“cs2 14877  ⟨“cs3 14878

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-8 2108  ax-9 2116  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1540  df-fal 1550  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-rab 3434  df-v 3480  df-dif 3966  df-un 3968  df-ss 3980  df-nul 4340  df-if 4532  df-sn 4632  df-pr 4634  df-op 4638  df-uni 4913  df-br 5149  df-iota 6516  df-fv 6571  df-ov 7434  df-s1 14631  df-s2 14884  df-s3 14885

This theorem is referenced by:  s4eqd  14901  s3eq2  14906  s3sndisj  15003  s3iunsndisj  15004  ragcgr  28730  perpneq  28737  isperp2  28738  isperp2d  28739  footexALT  28741  footexlem2  28743  foot  28745  perprag  28749  perpdragALT  28750  colperpexlem1  28753  lmiisolem  28819  hypcgrlem1  28822  hypcgrlem2  28823  trgcopyeu  28829  iscgra  28832  iscgra1  28833  iscgrad  28834  sacgr  28854  isleag  28870  isleagd  28871  iseqlg  28890