Theorem s8eqd 11269

Description: Equality theorem for a length 8 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)
s7eqd.6	⊢ (𝜑 → 𝐺 = 𝑇)
s8eqd.6	⊢ (𝜑 → 𝐻 = 𝑈)

Assertion

Ref	Expression
s8eqd	⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸𝐹𝐺𝐻”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅𝑆𝑇𝑈”⟩)

Proof of Theorem s8eqd

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		s6eqd.6	. . . 4 ⊢ (𝜑 → 𝐹 = 𝑆)
7		s7eqd.6	. . . 4 ⊢ (𝜑 → 𝐺 = 𝑇)
8	1, 2, 3, 4, 5, 6, 7	s7eqd 11268	. . 3 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸𝐹𝐺”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅𝑆𝑇”⟩)
9		s8eqd.6	. . . 4 ⊢ (𝜑 → 𝐻 = 𝑈)
10	9	s1eqd 11114	. . 3 ⊢ (𝜑 → ⟨“𝐻”⟩ = ⟨“𝑈”⟩)
11	8, 10	oveq12d 5987	. 2 ⊢ (𝜑 → (⟨“𝐴𝐵𝐶𝐷𝐸𝐹𝐺”⟩ ++ ⟨“𝐻”⟩) = (⟨“𝑁𝑂𝑃𝑄𝑅𝑆𝑇”⟩ ++ ⟨“𝑈”⟩))
12		df-s8 11255	. 2 ⊢ ⟨“𝐴𝐵𝐶𝐷𝐸𝐹𝐺𝐻”⟩ = (⟨“𝐴𝐵𝐶𝐷𝐸𝐹𝐺”⟩ ++ ⟨“𝐻”⟩)
13		df-s8 11255	. 2 ⊢ ⟨“𝑁𝑂𝑃𝑄𝑅𝑆𝑇𝑈”⟩ = (⟨“𝑁𝑂𝑃𝑄𝑅𝑆𝑇”⟩ ++ ⟨“𝑈”⟩)
14	11, 12, 13	3eqtr4g 2265	1 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸𝐹𝐺𝐻”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅𝑆𝑇𝑈”⟩)

Syntax hints:   → wi 4   = wceq 1373  (class class class)co 5969   ++ cconcat 11086  ⟨“cs1 11109  ⟨“cs7 11247  ⟨“cs8 11248

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2189

This theorem depends on definitions:  df-bi 117  df-3an 983  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2194  df-cleq 2200  df-clel 2203  df-nfc 2339  df-rex 2492  df-v 2779  df-un 3179  df-sn 3650  df-pr 3651  df-op 3653  df-uni 3866  df-br 4061  df-iota 5252  df-fv 5299  df-ov 5972  df-s1 11110  df-s2 11249  df-s3 11250  df-s4 11251  df-s5 11252  df-s6 11253  df-s7 11254  df-s8 11255

This theorem is referenced by: (None)