Theorem s5eqd 13462

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 13461	. . 3 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷”⟩ = ⟨“𝑁𝑂𝑃𝑄”⟩)
6		s5eqd.5	. . . 4 ⊢ (𝜑 → 𝐸 = 𝑅)
7	6	s1eqd 13234	. . 3 ⊢ (𝜑 → ⟨“𝐸”⟩ = ⟨“𝑅”⟩)
8	5, 7	oveq12d 6567	. 2 ⊢ (𝜑 → (⟨“𝐴𝐵𝐶𝐷”⟩ ++ ⟨“𝐸”⟩) = (⟨“𝑁𝑂𝑃𝑄”⟩ ++ ⟨“𝑅”⟩))
9		df-s5 13447	. 2 ⊢ ⟨“𝐴𝐵𝐶𝐷𝐸”⟩ = (⟨“𝐴𝐵𝐶𝐷”⟩ ++ ⟨“𝐸”⟩)
10		df-s5 13447	. 2 ⊢ ⟨“𝑁𝑂𝑃𝑄𝑅”⟩ = (⟨“𝑁𝑂𝑃𝑄”⟩ ++ ⟨“𝑅”⟩)
11	8, 9, 10	3eqtr4g 2669	1 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅”⟩)

Syntax hints:   → wi 4   = wceq 1475  (class class class)co 6549   ++ cconcat 13148  ⟨“cs1 13149  ⟨“cs4 13439  ⟨“cs5 13440

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-3an 1033  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-rex 2902  df-rab 2905  df-v 3175  df-dif 3543  df-un 3545  df-in 3547  df-ss 3554  df-nul 3875  df-if 4037  df-sn 4126  df-pr 4128  df-op 4132  df-uni 4373  df-br 4584  df-iota 5768  df-fv 5812  df-ov 6552  df-s1 13157  df-s2 13444  df-s3 13445  df-s4 13446  df-s5 13447

This theorem is referenced by:  s6eqd  13463