Theorem s6eqd 11306

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 11305	. . 3 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅”⟩)
7		s6eqd.6	. . . 4 ⊢ (𝜑 → 𝐹 = 𝑆)
8	7	s1eqd 11153	. . 3 ⊢ (𝜑 → ⟨“𝐹”⟩ = ⟨“𝑆”⟩)
9	6, 8	oveq12d 6019	. 2 ⊢ (𝜑 → (⟨“𝐴𝐵𝐶𝐷𝐸”⟩ ++ ⟨“𝐹”⟩) = (⟨“𝑁𝑂𝑃𝑄𝑅”⟩ ++ ⟨“𝑆”⟩))
10		df-s6 11292	. 2 ⊢ ⟨“𝐴𝐵𝐶𝐷𝐸𝐹”⟩ = (⟨“𝐴𝐵𝐶𝐷𝐸”⟩ ++ ⟨“𝐹”⟩)
11		df-s6 11292	. 2 ⊢ ⟨“𝑁𝑂𝑃𝑄𝑅𝑆”⟩ = (⟨“𝑁𝑂𝑃𝑄𝑅”⟩ ++ ⟨“𝑆”⟩)
12	9, 10, 11	3eqtr4g 2287	1 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸𝐹”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅𝑆”⟩)

Syntax hints:   → wi 4   = wceq 1395  (class class class)co 6001   ++ cconcat 11125  ⟨“cs1 11148  ⟨“cs5 11284  ⟨“cs6 11285

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 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-3an 1004  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-rex 2514  df-v 2801  df-un 3201  df-sn 3672  df-pr 3673  df-op 3675  df-uni 3889  df-br 4084  df-iota 5278  df-fv 5326  df-ov 6004  df-s1 11149  df-s2 11288  df-s3 11289  df-s4 11290  df-s5 11291  df-s6 11292

This theorem is referenced by:  s7eqd  11307