Theorem s4eqd 14928

Description: Equality theorem for a length 4 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)

Assertion

Ref	Expression
s4eqd	⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷”⟩ = ⟨“𝑁𝑂𝑃𝑄”⟩)

Proof of Theorem s4eqd

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	1, 2, 3	s3eqd 14927	. . 3 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶”⟩ = ⟨“𝑁𝑂𝑃”⟩)
5		s4eqd.4	. . . 4 ⊢ (𝜑 → 𝐷 = 𝑄)
6	5	s1eqd 14660	. . 3 ⊢ (𝜑 → ⟨“𝐷”⟩ = ⟨“𝑄”⟩)
7	4, 6	oveq12d 7437	. 2 ⊢ (𝜑 → (⟨“𝐴𝐵𝐶”⟩ ++ ⟨“𝐷”⟩) = (⟨“𝑁𝑂𝑃”⟩ ++ ⟨“𝑄”⟩))
8		df-s4 14913	. 2 ⊢ ⟨“𝐴𝐵𝐶𝐷”⟩ = (⟨“𝐴𝐵𝐶”⟩ ++ ⟨“𝐷”⟩)
9		df-s4 14913	. 2 ⊢ ⟨“𝑁𝑂𝑃𝑄”⟩ = (⟨“𝑁𝑂𝑃”⟩ ++ ⟨“𝑄”⟩)
10	7, 8, 9	3eqtr4g 2825	1 ⊢ (𝜑 → ⟨“𝐴𝐵𝐶𝐷”⟩ = ⟨“𝑁𝑂𝑃𝑄”⟩)

This proof depends on syntax axioms:   → wi 4   = wceq 1570  (class class class)co 7419   ++ cconcat 14627  ⟨“cs1 14654  ⟨“cs3 14905  ⟨“cs4 14906

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-uni 4875  df-br 5112  df-iota 6496  df-fv 6548  df-ov 7422  df-s1 14655  df-s2 14911  df-s3 14912  df-s4 14913

This theorem is used by:  s5eqd  14929