Theorem s5eqd 14823

Description: Equality theorem for a length 5 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)

Assertion

Ref	Expression
s5eqd	⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅”⟩)

Proof of Theorem s5eqd

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	1, 2, 3, 4	s4eqd 14822	. . 3 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷”⟩ = ⟨“𝑁𝑂𝑃𝑄”⟩)
6		s5eqd.5	. . . 4 ⊢ (𝜑 → 𝐸 = 𝑅)
7	6	s1eqd 14559	. . 3 ⊢ (𝜑 → ⟨“𝐸”⟩ = ⟨“𝑅”⟩)
8	5, 7	oveq12d 7377	. 2 ⊢ (𝜑 → (⟨“𝐴𝐵𝐶𝐷”⟩ ++ ⟨“𝐸”⟩) = (⟨“𝑁𝑂𝑃𝑄”⟩ ++ ⟨“𝑅”⟩))
9		df-s5 14808	. 2 ⊢ ⟨“𝐴𝐵𝐶𝐷𝐸”⟩ = (⟨“𝐴𝐵𝐶𝐷”⟩ ++ ⟨“𝐸”⟩)
10		df-s5 14808	. 2 ⊢ ⟨“𝑁𝑂𝑃𝑄𝑅”⟩ = (⟨“𝑁𝑂𝑃𝑄”⟩ ++ ⟨“𝑅”⟩)
11	8, 9, 10	3eqtr4g 2801	1 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅”⟩)

Syntax hints:   → wi 4   = wceq 1548  (class class class)co 7359   ++ cconcat 14527  ⟨“cs1 14553  ⟨“cs4 14800  ⟨“cs5 14801

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1975  ax-7 2016  ax-8 2123  ax-9 2131  ax-ext 2713

This theorem depends on definitions:  df-bi 209  df-an 398  df-or 855  df-3an 1095  df-tru 1551  df-fal 1561  df-ex 1788  df-sb 2075  df-clab 2720  df-cleq 2733  df-clel 2816  df-rab 3394  df-v 3435  df-dif 3887  df-un 3889  df-ss 3901  df-nul 4264  df-if 4457  df-sn 4558  df-pr 4560  df-op 4564  df-uni 4841  df-br 5075  df-iota 6444  df-fv 6496  df-ov 7362  df-s1 14554  df-s2 14805  df-s3 14806  df-s4 14807  df-s5 14808

This theorem is referenced by:  s6eqd  14824