Theorem s3eqd

Description: Equality theorem for a length 3 word. (Contributed by Mario Carneiro, 27-Feb-2016)

	Ref	Expression
Hypotheses	s2eqd.1	⊢ ( 𝜑 → 𝐴 = 𝑁 )
	s2eqd.2	⊢ ( 𝜑 → 𝐵 = 𝑂 )
	s3eqd.3	⊢ ( 𝜑 → 𝐶 = 𝑃 )
Assertion	s3eqd	⊢ ( 𝜑 → ⟨“ 𝐴 𝐵 𝐶 ”⟩ = ⟨“ 𝑁 𝑂 𝑃 ”⟩ )

Proof

Step	Hyp	Ref	Expression
1		s2eqd.1	⊢ ( 𝜑 → 𝐴 = 𝑁 )
2		s2eqd.2	⊢ ( 𝜑 → 𝐵 = 𝑂 )
3		s3eqd.3	⊢ ( 𝜑 → 𝐶 = 𝑃 )
4	1 2	s2eqd	⊢ ( 𝜑 → ⟨“ 𝐴 𝐵 ”⟩ = ⟨“ 𝑁 𝑂 ”⟩ )
5	3	s1eqd	⊢ ( 𝜑 → ⟨“ 𝐶 ”⟩ = ⟨“ 𝑃 ”⟩ )
6	4 5	oveq12d	⊢ ( 𝜑 → ( ⟨“ 𝐴 𝐵 ”⟩ ++ ⟨“ 𝐶 ”⟩ ) = ( ⟨“ 𝑁 𝑂 ”⟩ ++ ⟨“ 𝑃 ”⟩ ) )
7		df-s3	⊢ ⟨“ 𝐴 𝐵 𝐶 ”⟩ = ( ⟨“ 𝐴 𝐵 ”⟩ ++ ⟨“ 𝐶 ”⟩ )
8		df-s3	⊢ ⟨“ 𝑁 𝑂 𝑃 ”⟩ = ( ⟨“ 𝑁 𝑂 ”⟩ ++ ⟨“ 𝑃 ”⟩ )
9	6 7 8	3eqtr4g	⊢ ( 𝜑 → ⟨“ 𝐴 𝐵 𝐶 ”⟩ = ⟨“ 𝑁 𝑂 𝑃 ”⟩ )