Theorem ndmovcom

Description: Any operation is commutative outside its domain. (Contributed by NM, 24-Aug-1995)

	Ref	Expression
Hypothesis	ndmov.1	⊢ dom 𝐹 = ( 𝑆 × 𝑆 )
Assertion	ndmovcom	⊢ ( ¬ ( 𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆 ) → ( 𝐴 𝐹 𝐵 ) = ( 𝐵 𝐹 𝐴 ) )

Proof

Step	Hyp	Ref	Expression
1		ndmov.1	⊢ dom 𝐹 = ( 𝑆 × 𝑆 )
2	1	ndmov	⊢ ( ¬ ( 𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆 ) → ( 𝐴 𝐹 𝐵 ) = ∅ )
3		ancom	⊢ ( ( 𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆 ) ↔ ( 𝐵 ∈ 𝑆 ∧ 𝐴 ∈ 𝑆 ) )
4	1	ndmov	⊢ ( ¬ ( 𝐵 ∈ 𝑆 ∧ 𝐴 ∈ 𝑆 ) → ( 𝐵 𝐹 𝐴 ) = ∅ )
5	3 4	sylnbi	⊢ ( ¬ ( 𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆 ) → ( 𝐵 𝐹 𝐴 ) = ∅ )
6	2 5	eqtr4d	⊢ ( ¬ ( 𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆 ) → ( 𝐴 𝐹 𝐵 ) = ( 𝐵 𝐹 𝐴 ) )