Metamath Proof Explorer


Theorem pfxnndmnd

Description: The value of a prefix operation for out-of-domain arguments. (This is due to our definition of function values for out-of-domain arguments, see ndmfv ). (Contributed by AV, 3-Dec-2022) (New usage is discouraged.)

Ref Expression
Assertion pfxnndmnd ( ¬ ( 𝑆 ∈ V ∧ 𝐿 ∈ ℕ0 ) → ( 𝑆 prefix 𝐿 ) = ∅ )

Proof

Step Hyp Ref Expression
1 df-pfx prefix = ( 𝑠 ∈ V , 𝑙 ∈ ℕ0 ↦ ( 𝑠 substr ⟨ 0 , 𝑙 ⟩ ) )
2 1 mpondm0 ( ¬ ( 𝑆 ∈ V ∧ 𝐿 ∈ ℕ0 ) → ( 𝑆 prefix 𝐿 ) = ∅ )