Theorem s3eqd 14867

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 14866	. . 3 ⊢ (𝜑 → ⟨“𝐴𝐵”⟩ = ⟨“𝑁𝑂”⟩)
4		s3eqd.3	. . . 4 ⊢ (𝜑 → 𝐶 = 𝑃)
5	4	s1eqd 14605	. . 3 ⊢ (𝜑 → ⟨“𝐶”⟩ = ⟨“𝑃”⟩)
6	3, 5	oveq12d 7403	. 2 ⊢ (𝜑 → (⟨“𝐴𝐵”⟩ ++ ⟨“𝐶”⟩) = (⟨“𝑁𝑂”⟩ ++ ⟨“𝑃”⟩))
7		df-s3 14852	. 2 ⊢ ⟨“𝐴𝐵𝐶”⟩ = (⟨“𝐴𝐵”⟩ ++ ⟨“𝐶”⟩)
8		df-s3 14852	. 2 ⊢ ⟨“𝑁𝑂𝑃”⟩ = (⟨“𝑁𝑂”⟩ ++ ⟨“𝑃”⟩)
9	6, 7, 8	3eqtr4g 2816	1 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶”⟩ = ⟨“𝑁𝑂𝑃”⟩)

Syntax hints:   → wi 4   = wceq 1554  (class class class)co 7385   ++ cconcat 14573  ⟨“cs1 14599  ⟨“cs2 14844  ⟨“cs3 14845

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1809  ax-4 1823  ax-5 1924  ax-6 1981  ax-7 2022  ax-8 2138  ax-9 2146  ax-ext 2728

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 857  df-3an 1097  df-tru 1557  df-fal 1567  df-ex 1794  df-sb 2085  df-clab 2735  df-cleq 2748  df-clel 2831  df-rab 3409  df-v 3450  df-dif 3902  df-un 3904  df-ss 3916  df-nul 4281  df-if 4475  df-sn 4577  df-pr 4579  df-op 4583  df-uni 4860  df-br 5095  df-iota 6466  df-fv 6518  df-ov 7388  df-s1 14600  df-s2 14851  df-s3 14852

This theorem is referenced by:  s4eqd  14868  s3eq2  14873  s3sndisj  14970  s3iunsndisj  14971  ragcgr  28846  perpneq  28853  isperp2  28854  isperp2d  28855  footexALT  28857  footexlem2  28859  foot  28861  perprag  28865  perpdragALT  28866  colperpexlem1  28869  lmiisolem  28935  hypcgrlem1  28938  hypcgrlem2  28939  trgcopyeu  28945  iscgra  28948  iscgra1  28949  iscgrad  28950  sacgr  28970  isleag  28986  isleagd  28987  iseqlg  29006