Theorem s3eqd 14817

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 14816	. . 3 ⊢ (𝜑 → ⟨“𝐴𝐵”⟩ = ⟨“𝑁𝑂”⟩)
4		s3eqd.3	. . . 4 ⊢ (𝜑 → 𝐶 = 𝑃)
5	4	s1eqd 14555	. . 3 ⊢ (𝜑 → ⟨“𝐶”⟩ = ⟨“𝑃”⟩)
6	3, 5	oveq12d 7378	. 2 ⊢ (𝜑 → (⟨“𝐴𝐵”⟩ ++ ⟨“𝐶”⟩) = (⟨“𝑁𝑂”⟩ ++ ⟨“𝑃”⟩))
7		df-s3 14802	. 2 ⊢ ⟨“𝐴𝐵𝐶”⟩ = (⟨“𝐴𝐵”⟩ ++ ⟨“𝐶”⟩)
8		df-s3 14802	. 2 ⊢ ⟨“𝑁𝑂𝑃”⟩ = (⟨“𝑁𝑂”⟩ ++ ⟨“𝑃”⟩)
9	6, 7, 8	3eqtr4g 2797	1 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶”⟩ = ⟨“𝑁𝑂𝑃”⟩)

Syntax hints:   → wi 4   = wceq 1542  (class class class)co 7360   ++ cconcat 14523  ⟨“cs1 14549  ⟨“cs2 14794  ⟨“cs3 14795

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-rab 3391  df-v 3432  df-dif 3893  df-un 3895  df-ss 3907  df-nul 4275  df-if 4468  df-sn 4569  df-pr 4571  df-op 4575  df-uni 4852  df-br 5087  df-iota 6448  df-fv 6500  df-ov 7363  df-s1 14550  df-s2 14801  df-s3 14802

This theorem is referenced by:  s4eqd  14818  s3eq2  14823  s3sndisj  14920  s3iunsndisj  14921  ragcgr  28789  perpneq  28796  isperp2  28797  isperp2d  28798  footexALT  28800  footexlem2  28802  foot  28804  perprag  28808  perpdragALT  28809  colperpexlem1  28812  lmiisolem  28878  hypcgrlem1  28881  hypcgrlem2  28882  trgcopyeu  28888  iscgra  28891  iscgra1  28892  iscgrad  28893  sacgr  28913  isleag  28929  isleagd  28930  iseqlg  28949