Theorem s5eqd 11545

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 11544	. . 3 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷”⟩ = ⟨“𝑁𝑂𝑃𝑄”⟩)
6		s5eqd.5	. . . 4 ⊢ (𝜑 → 𝐸 = 𝑅)
7	6	s1eqd 11388	. . 3 ⊢ (𝜑 → ⟨“𝐸”⟩ = ⟨“𝑅”⟩)
8	5, 7	oveq12d 6103	. 2 ⊢ (𝜑 → (⟨“𝐴𝐵𝐶𝐷”⟩ ++ ⟨“𝐸”⟩) = (⟨“𝑁𝑂𝑃𝑄”⟩ ++ ⟨“𝑅”⟩))
9		df-s5 11531	. 2 ⊢ ⟨“𝐴𝐵𝐶𝐷𝐸”⟩ = (⟨“𝐴𝐵𝐶𝐷”⟩ ++ ⟨“𝐸”⟩)
10		df-s5 11531	. 2 ⊢ ⟨“𝑁𝑂𝑃𝑄𝑅”⟩ = (⟨“𝑁𝑂𝑃𝑄”⟩ ++ ⟨“𝑅”⟩)
11	8, 9, 10	3eqtr4g 2296	1 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷𝐸”⟩ = ⟨“𝑁𝑂𝑃𝑄𝑅”⟩)

This proof depends on syntax axioms:   → wi 4   = wceq 1402  (class class class)co 6085   ++ cconcat 11358  ⟨“cs1 11383  ⟨“cs4 11523  ⟨“cs5 11524

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-rex 2534  df-v 2823  df-un 3224  df-sn 3715  df-pr 3716  df-op 3718  df-uni 3936  df-br 4131  df-iota 5337  df-fv 5385  df-ov 6088  df-s1 11384  df-s2 11528  df-s3 11529  df-s4 11530  df-s5 11531

This theorem is used by:  s6eqd  11546