Theorem s4eqd

Description: Equality theorem for a length 4 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 )
	s4eqd.4	⊢ ( 𝜑 → 𝐷 = 𝑄 )
Assertion	s4eqd	⊢ ( 𝜑 → ⟨“ 𝐴 𝐵 𝐶 𝐷 ”⟩ = ⟨“ 𝑁 𝑂 𝑃 𝑄 ”⟩ )

Proof

Step	Hyp	Ref	Expression
1		s2eqd.1	⊢ ( 𝜑 → 𝐴 = 𝑁 )
2		s2eqd.2	⊢ ( 𝜑 → 𝐵 = 𝑂 )
3		s3eqd.3	⊢ ( 𝜑 → 𝐶 = 𝑃 )
4		s4eqd.4	⊢ ( 𝜑 → 𝐷 = 𝑄 )
5	1 2 3	s3eqd	⊢ ( 𝜑 → ⟨“ 𝐴 𝐵 𝐶 ”⟩ = ⟨“ 𝑁 𝑂 𝑃 ”⟩ )
6	4	s1eqd	⊢ ( 𝜑 → ⟨“ 𝐷 ”⟩ = ⟨“ 𝑄 ”⟩ )
7	5 6	oveq12d	⊢ ( 𝜑 → ( ⟨“ 𝐴 𝐵 𝐶 ”⟩ ++ ⟨“ 𝐷 ”⟩ ) = ( ⟨“ 𝑁 𝑂 𝑃 ”⟩ ++ ⟨“ 𝑄 ”⟩ ) )
8		df-s4	⊢ ⟨“ 𝐴 𝐵 𝐶 𝐷 ”⟩ = ( ⟨“ 𝐴 𝐵 𝐶 ”⟩ ++ ⟨“ 𝐷 ”⟩ )
9		df-s4	⊢ ⟨“ 𝑁 𝑂 𝑃 𝑄 ”⟩ = ( ⟨“ 𝑁 𝑂 𝑃 ”⟩ ++ ⟨“ 𝑄 ”⟩ )
10	7 8 9	3eqtr4g	⊢ ( 𝜑 → ⟨“ 𝐴 𝐵 𝐶 𝐷 ”⟩ = ⟨“ 𝑁 𝑂 𝑃 𝑄 ”⟩ )