Metamath Proof Explorer


Theorem ndmovcl

Description: The closure of an operation outside its domain, when the domain includes the empty set. This technical lemma can make the operation more convenient to work in some cases. It is dependent on our particular definitions of operation value, function value, and ordered pair. (Contributed by NM, 24-Sep-2004)

Ref Expression
Hypotheses ndmov.1 dom 𝐹 = ( 𝑆 × 𝑆 )
ndmovcl.2 ( ( 𝐴𝑆𝐵𝑆 ) → ( 𝐴 𝐹 𝐵 ) ∈ 𝑆 )
ndmovcl.3 ∅ ∈ 𝑆
Assertion ndmovcl ( 𝐴 𝐹 𝐵 ) ∈ 𝑆

Proof

Step Hyp Ref Expression
1 ndmov.1 dom 𝐹 = ( 𝑆 × 𝑆 )
2 ndmovcl.2 ( ( 𝐴𝑆𝐵𝑆 ) → ( 𝐴 𝐹 𝐵 ) ∈ 𝑆 )
3 ndmovcl.3 ∅ ∈ 𝑆
4 1 ndmov ( ¬ ( 𝐴𝑆𝐵𝑆 ) → ( 𝐴 𝐹 𝐵 ) = ∅ )
5 4 3 eqeltrdi ( ¬ ( 𝐴𝑆𝐵𝑆 ) → ( 𝐴 𝐹 𝐵 ) ∈ 𝑆 )
6 2 5 pm2.61i ( 𝐴 𝐹 𝐵 ) ∈ 𝑆