Theorem s6eqd 14223

Description: Equality theorem for a length 6 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)
s4eqd.4	⊢ (𝜑 → 𝐷 = 𝑄)
s5eqd.5	⊢ (𝜑 → 𝐸 = 𝑅)
s6eqd.6	⊢ (𝜑 → 𝐹 = 𝑆)

Assertion

Ref	Expression
s6eqd	⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸𝐹”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅𝑆”⟩)

Proof of Theorem s6eqd

Step	Hyp	Ref	Expression
1		s2eqd.1	. . . 4 ⊢ (𝜑 → 𝐴 = 𝑁)
2		s2eqd.2	. . . 4 ⊢ (𝜑 → 𝐵 = 𝑂)
3		s3eqd.3	. . . 4 ⊢ (𝜑 → 𝐶 = 𝑃)
4		s4eqd.4	. . . 4 ⊢ (𝜑 → 𝐷 = 𝑄)
5		s5eqd.5	. . . 4 ⊢ (𝜑 → 𝐸 = 𝑅)
6	1, 2, 3, 4, 5	s5eqd 14222	. . 3 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅”⟩)
7		s6eqd.6	. . . 4 ⊢ (𝜑 → 𝐹 = 𝑆)
8	7	s1eqd 13949	. . 3 ⊢ (𝜑 → ⟨“𝐹”⟩ = ⟨“𝑆”⟩)
9	6, 8	oveq12d 7168	. 2 ⊢ (𝜑 → (⟨“𝐴𝐵𝐶𝐷𝐸”⟩ ++ ⟨“𝐹”⟩) = (⟨“𝑁𝑂𝑃𝑄𝑅”⟩ ++ ⟨“𝑆”⟩))
10		df-s6 14208	. 2 ⊢ ⟨“𝐴𝐵𝐶𝐷𝐸𝐹”⟩ = (⟨“𝐴𝐵𝐶𝐷𝐸”⟩ ++ ⟨“𝐹”⟩)
11		df-s6 14208	. 2 ⊢ ⟨“𝑁𝑂𝑃𝑄𝑅𝑆”⟩ = (⟨“𝑁𝑂𝑃𝑄𝑅”⟩ ++ ⟨“𝑆”⟩)
12	9, 10, 11	3eqtr4g 2881	1 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸𝐹”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅𝑆”⟩)

Syntax hints:   → wi 4   = wceq 1533  (class class class)co 7150   ++ cconcat 13916  ⟨“cs1 13943  ⟨“cs5 14200  ⟨“cs6 14201

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 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-rex 3144  df-rab 3147  df-v 3496  df-dif 3938  df-un 3940  df-in 3942  df-ss 3951  df-nul 4291  df-if 4467  df-sn 4561  df-pr 4563  df-op 4567  df-uni 4832  df-br 5059  df-iota 6308  df-fv 6357  df-ov 7153  df-s1 13944  df-s2 14204  df-s3 14205  df-s4 14206  df-s5 14207  df-s6 14208

This theorem is referenced by:  s7eqd  14224