Metamath Proof Explorer


Theorem ndmovdistr

Description: Any operation is distributive outside its domain, if the domain doesn't contain the empty set. (Contributed by NM, 24-Aug-1995)

Ref Expression
Hypotheses ndmov.1 ⊢ dom ⁡ F = S × S
ndmov.5 ⊢ ¬ ∅ ∈ S
ndmov.6 ⊢ dom ⁡ G = S × S
Assertion ndmovdistr ⊢ ¬ A ∈ S ∧ B ∈ S ∧ C ∈ S → A G B F C = A G B F A G C

Proof

Step Hyp Ref Expression
1 ndmov.1 ⊢ dom ⁡ F = S × S
2 ndmov.5 ⊢ ¬ ∅ ∈ S
3 ndmov.6 ⊢ dom ⁡ G = S × S
4 1 2 ndmovrcl ⊢ B F C ∈ S → B ∈ S ∧ C ∈ S
5 4 anim2i ⊢ A ∈ S ∧ B F C ∈ S → A ∈ S ∧ B ∈ S ∧ C ∈ S
6 3anass ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ↔ A ∈ S ∧ B ∈ S ∧ C ∈ S
7 5 6 sylibr ⊢ A ∈ S ∧ B F C ∈ S → A ∈ S ∧ B ∈ S ∧ C ∈ S
8 3 ndmov ⊢ ¬ A ∈ S ∧ B F C ∈ S → A G B F C = ∅
9 7 8 nsyl5 ⊢ ¬ A ∈ S ∧ B ∈ S ∧ C ∈ S → A G B F C = ∅
10 3 2 ndmovrcl ⊢ A G B ∈ S → A ∈ S ∧ B ∈ S
11 3 2 ndmovrcl ⊢ A G C ∈ S → A ∈ S ∧ C ∈ S
12 10 11 anim12i ⊢ A G B ∈ S ∧ A G C ∈ S → A ∈ S ∧ B ∈ S ∧ A ∈ S ∧ C ∈ S
13 anandi3 ⊢ A ∈ S ∧ B ∈ S ∧ C ∈ S ↔ A ∈ S ∧ B ∈ S ∧ A ∈ S ∧ C ∈ S
14 12 13 sylibr ⊢ A G B ∈ S ∧ A G C ∈ S → A ∈ S ∧ B ∈ S ∧ C ∈ S
15 1 ndmov ⊢ ¬ A G B ∈ S ∧ A G C ∈ S → A G B F A G C = ∅
16 14 15 nsyl5 ⊢ ¬ A ∈ S ∧ B ∈ S ∧ C ∈ S → A G B F A G C = ∅
17 9 16 eqtr4d ⊢ ¬ A ∈ S ∧ B ∈ S ∧ C ∈ S → A G B F C = A G B F A G C