Theorem s5eqd 14816

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 14815	. . 3 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷”⟩ = ⟨“𝑁𝑂𝑃𝑄”⟩)
6		s5eqd.5	. . . 4 ⊢ (𝜑 → 𝐸 = 𝑅)
7	6	s1eqd 14550	. . 3 ⊢ (𝜑 → ⟨“𝐸”⟩ = ⟨“𝑅”⟩)
8	5, 7	oveq12d 7426	. 2 ⊢ (𝜑 → (⟨“𝐴𝐵𝐶𝐷”⟩ ++ ⟨“𝐸”⟩) = (⟨“𝑁𝑂𝑃𝑄”⟩ ++ ⟨“𝑅”⟩))
9		df-s5 14801	. 2 ⊢ ⟨“𝐴𝐵𝐶𝐷𝐸”⟩ = (⟨“𝐴𝐵𝐶𝐷”⟩ ++ ⟨“𝐸”⟩)
10		df-s5 14801	. 2 ⊢ ⟨“𝑁𝑂𝑃𝑄𝑅”⟩ = (⟨“𝑁𝑂𝑃𝑄”⟩ ++ ⟨“𝑅”⟩)
11	8, 9, 10	3eqtr4g 2797	1 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅”⟩)

Syntax hints:   → wi 4   = wceq 1541  (class class class)co 7408   ++ cconcat 14519  ⟨“cs1 14544  ⟨“cs4 14793  ⟨“cs5 14794

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2703

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-3an 1089  df-tru 1544  df-fal 1554  df-ex 1782  df-sb 2068  df-clab 2710  df-cleq 2724  df-clel 2810  df-rab 3433  df-v 3476  df-dif 3951  df-un 3953  df-in 3955  df-ss 3965  df-nul 4323  df-if 4529  df-sn 4629  df-pr 4631  df-op 4635  df-uni 4909  df-br 5149  df-iota 6495  df-fv 6551  df-ov 7411  df-s1 14545  df-s2 14798  df-s3 14799  df-s4 14800  df-s5 14801

This theorem is referenced by:  s6eqd  14817