Theorem s3eqd 14830

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 14829	. . 3 ⊢ (𝜑 → ⟨“𝐴𝐵”⟩ = ⟨“𝑁𝑂”⟩)
4		s3eqd.3	. . . 4 ⊢ (𝜑 → 𝐶 = 𝑃)
5	4	s1eqd 14566	. . 3 ⊢ (𝜑 → ⟨“𝐶”⟩ = ⟨“𝑃”⟩)
6	3, 5	oveq12d 7405	. 2 ⊢ (𝜑 → (⟨“𝐴𝐵”⟩ ++ ⟨“𝐶”⟩) = (⟨“𝑁𝑂”⟩ ++ ⟨“𝑃”⟩))
7		df-s3 14815	. 2 ⊢ ⟨“𝐴𝐵𝐶”⟩ = (⟨“𝐴𝐵”⟩ ++ ⟨“𝐶”⟩)
8		df-s3 14815	. 2 ⊢ ⟨“𝑁𝑂𝑃”⟩ = (⟨“𝑁𝑂”⟩ ++ ⟨“𝑃”⟩)
9	6, 7, 8	3eqtr4g 2789	1 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶”⟩ = ⟨“𝑁𝑂𝑃”⟩)

Syntax hints:   → wi 4   = wceq 1540  (class class class)co 7387   ++ cconcat 14535  ⟨“cs1 14560  ⟨“cs2 14807  ⟨“cs3 14808

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2701

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-rab 3406  df-v 3449  df-dif 3917  df-un 3919  df-ss 3931  df-nul 4297  df-if 4489  df-sn 4590  df-pr 4592  df-op 4596  df-uni 4872  df-br 5108  df-iota 6464  df-fv 6519  df-ov 7390  df-s1 14561  df-s2 14814  df-s3 14815

This theorem is referenced by:  s4eqd  14831  s3eq2  14836  s3sndisj  14933  s3iunsndisj  14934  ragcgr  28634  perpneq  28641  isperp2  28642  isperp2d  28643  footexALT  28645  footexlem2  28647  foot  28649  perprag  28653  perpdragALT  28654  colperpexlem1  28657  lmiisolem  28723  hypcgrlem1  28726  hypcgrlem2  28727  trgcopyeu  28733  iscgra  28736  iscgra1  28737  iscgrad  28738  sacgr  28758  isleag  28774  isleagd  28775  iseqlg  28794