Theorem s6eqd 11529

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 11528	. . 3 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅”⟩)
7		s6eqd.6	. . . 4 ⊢ (𝜑 → 𝐹 = 𝑆)
8	7	s1eqd 11371	. . 3 ⊢ (𝜑 → ⟨“𝐹”⟩ = ⟨“𝑆”⟩)
9	6, 8	oveq12d 6097	. 2 ⊢ (𝜑 → (⟨“𝐴𝐵𝐶𝐷𝐸”⟩ ++ ⟨“𝐹”⟩) = (⟨“𝑁𝑂𝑃𝑄𝑅”⟩ ++ ⟨“𝑆”⟩))
10		df-s6 11515	. 2 ⊢ ⟨“𝐴𝐵𝐶𝐷𝐸𝐹”⟩ = (⟨“𝐴𝐵𝐶𝐷𝐸”⟩ ++ ⟨“𝐹”⟩)
11		df-s6 11515	. 2 ⊢ ⟨“𝑁𝑂𝑃𝑄𝑅𝑆”⟩ = (⟨“𝑁𝑂𝑃𝑄𝑅”⟩ ++ ⟨“𝑆”⟩)
12	9, 10, 11	3eqtr4g 2296	1 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸𝐹”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅𝑆”⟩)

Syntax hints:   → wi 4   = wceq 1402  (class class class)co 6079   ++ cconcat 11341  ⟨“cs1 11366  ⟨“cs5 11507  ⟨“cs6 11508

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 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-rex 2534  df-v 2823  df-un 3224  df-sn 3714  df-pr 3715  df-op 3717  df-uni 3934  df-br 4129  df-iota 5335  df-fv 5383  df-ov 6082  df-s1 11367  df-s2 11511  df-s3 11512  df-s4 11513  df-s5 11514  df-s6 11515

This theorem is referenced by:  s7eqd  11530