Metamath Proof Explorer


Theorem s5cld

Description: A length 5 string is a word. (Contributed by Mario Carneiro, 27-Feb-2016)

Ref Expression
Hypotheses s2cld.1 ⊢ φ → A ∈ X
s2cld.2 ⊢ φ → B ∈ X
s3cld.3 ⊢ φ → C ∈ X
s4cld.4 ⊢ φ → D ∈ X
s5cld.5 ⊢ φ → E ∈ X
Assertion s5cld ⊢ φ → ⟨“ ABCDE ”⟩ ∈ Word X

Proof

Step Hyp Ref Expression
1 s2cld.1 ⊢ φ → A ∈ X
2 s2cld.2 ⊢ φ → B ∈ X
3 s3cld.3 ⊢ φ → C ∈ X
4 s4cld.4 ⊢ φ → D ∈ X
5 s5cld.5 ⊢ φ → E ∈ X
6 df-s5 ⊢ ⟨“ ABCDE ”⟩ = ⟨“ ABCD ”⟩ ++ ⟨“ E ”⟩
7 1 2 3 4 s4cld ⊢ φ → ⟨“ ABCD ”⟩ ∈ Word X
8 6 7 5 cats1cld ⊢ φ → ⟨“ ABCDE ”⟩ ∈ Word X