Theorem s5eqd 14819

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 14818	. . 3 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷”⟩ = ⟨“𝑁𝑂𝑃𝑄”⟩)
6		s5eqd.5	. . . 4 ⊢ (𝜑 → 𝐸 = 𝑅)
7	6	s1eqd 14555	. . 3 ⊢ (𝜑 → ⟨“𝐸”⟩ = ⟨“𝑅”⟩)
8	5, 7	oveq12d 7374	. 2 ⊢ (𝜑 → (⟨“𝐴𝐵𝐶𝐷”⟩ ++ ⟨“𝐸”⟩) = (⟨“𝑁𝑂𝑃𝑄”⟩ ++ ⟨“𝑅”⟩))
9		df-s5 14804	. 2 ⊢ ⟨“𝐴𝐵𝐶𝐷𝐸”⟩ = (⟨“𝐴𝐵𝐶𝐷”⟩ ++ ⟨“𝐸”⟩)
10		df-s5 14804	. 2 ⊢ ⟨“𝑁𝑂𝑃𝑄𝑅”⟩ = (⟨“𝑁𝑂𝑃𝑄”⟩ ++ ⟨“𝑅”⟩)
11	8, 9, 10	3eqtr4g 2799	1 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅”⟩)

Syntax hints:   → wi 4   = wceq 1547  (class class class)co 7356   ++ cconcat 14523  ⟨“cs1 14549  ⟨“cs4 14796  ⟨“cs5 14797

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2711

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-3an 1094  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-clab 2718  df-cleq 2731  df-clel 2814  df-rab 3392  df-v 3433  df-dif 3886  df-un 3888  df-ss 3900  df-nul 4262  df-if 4455  df-sn 4556  df-pr 4558  df-op 4562  df-uni 4839  df-br 5073  df-iota 6441  df-fv 6493  df-ov 7359  df-s1 14550  df-s2 14801  df-s3 14802  df-s4 14803  df-s5 14804

This theorem is referenced by:  s6eqd  14820