Metamath Proof Explorer


Theorem s2eqd

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

Ref Expression
Hypotheses s2eqd.1 ( 𝜑𝐴 = 𝑁 )
s2eqd.2 ( 𝜑𝐵 = 𝑂 )
Assertion s2eqd ( 𝜑 → ⟨“ 𝐴 𝐵 ”⟩ = ⟨“ 𝑁 𝑂 ”⟩ )

Proof

Step Hyp Ref Expression
1 s2eqd.1 ( 𝜑𝐴 = 𝑁 )
2 s2eqd.2 ( 𝜑𝐵 = 𝑂 )
3 1 s1eqd ( 𝜑 → ⟨“ 𝐴 ”⟩ = ⟨“ 𝑁 ”⟩ )
4 2 s1eqd ( 𝜑 → ⟨“ 𝐵 ”⟩ = ⟨“ 𝑂 ”⟩ )
5 3 4 oveq12d ( 𝜑 → ( ⟨“ 𝐴 ”⟩ ++ ⟨“ 𝐵 ”⟩ ) = ( ⟨“ 𝑁 ”⟩ ++ ⟨“ 𝑂 ”⟩ ) )
6 df-s2 ⟨“ 𝐴 𝐵 ”⟩ = ( ⟨“ 𝐴 ”⟩ ++ ⟨“ 𝐵 ”⟩ )
7 df-s2 ⟨“ 𝑁 𝑂 ”⟩ = ( ⟨“ 𝑁 ”⟩ ++ ⟨“ 𝑂 ”⟩ )
8 5 6 7 3eqtr4g ( 𝜑 → ⟨“ 𝐴 𝐵 ”⟩ = ⟨“ 𝑁 𝑂 ”⟩ )