Theorem s7eqd 15033

Description: Equality theorem for a length 7 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)

Assertion

Ref	Expression
s7eqd	⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸𝐹𝐺”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅𝑆𝑇”⟩)

Proof of Theorem s7eqd

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	1, 2, 3, 4, 5, 6	s6eqd 15032	. . 3 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸𝐹”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅𝑆”⟩)
8		s7eqd.6	. . . 4 ⊢ (𝜑 → 𝐺 = 𝑇)
9	8	s1eqd 14761	. . 3 ⊢ (𝜑 → ⟨“𝐺”⟩ = ⟨“𝑇”⟩)
10	7, 9	oveq12d 7448	. 2 ⊢ (𝜑 → (⟨“𝐴𝐵𝐶𝐷𝐸𝐹”⟩ ++ ⟨“𝐺”⟩) = (⟨“𝑁𝑂𝑃𝑄𝑅𝑆”⟩ ++ ⟨“𝑇”⟩))
11		df-s7 15018	. 2 ⊢ ⟨“𝐴𝐵𝐶𝐷𝐸𝐹𝐺”⟩ = (⟨“𝐴𝐵𝐶𝐷𝐸𝐹”⟩ ++ ⟨“𝐺”⟩)
12		df-s7 15018	. 2 ⊢ ⟨“𝑁𝑂𝑃𝑄𝑅𝑆𝑇”⟩ = (⟨“𝑁𝑂𝑃𝑄𝑅𝑆”⟩ ++ ⟨“𝑇”⟩)
13	10, 11, 12	3eqtr4g 2823	1 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸𝐹𝐺”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅𝑆𝑇”⟩)

This proof depends on syntax axioms:   → wi 4   = wceq 1569  (class class class)co 7430   ++ cconcat 14728  ⟨“cs1 14755  ⟨“cs6 15010  ⟨“cs7 15011

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-uni 4871  df-br 5107  df-iota 6501  df-fv 6555  df-ov 7433  df-s1 14756  df-s2 15013  df-s3 15014  df-s4 15015  df-s5 15016  df-s6 15017  df-s7 15018

This theorem is used by:  s8eqd  15034