Metamath Proof Explorer


Theorem coeq0i

Description: coeq0 but without explicitly introducing domain and range symbols. (Contributed by Stefan O'Rear, 16-Oct-2014)

Ref Expression
Assertion coeq0i ⊢ A : C ⟶ D ∧ B : E ⟶ F ∧ C ∩ F = ∅ → A ∘ B = ∅

Proof

Step Hyp Ref Expression
1 frn ⊢ B : E ⟶ F → ran ⁡ B ⊆ F
2 1 3ad2ant2 ⊢ A : C ⟶ D ∧ B : E ⟶ F ∧ C ∩ F = ∅ → ran ⁡ B ⊆ F
3 sslin ⊢ ran ⁡ B ⊆ F → dom ⁡ A ∩ ran ⁡ B ⊆ dom ⁡ A ∩ F
4 2 3 syl ⊢ A : C ⟶ D ∧ B : E ⟶ F ∧ C ∩ F = ∅ → dom ⁡ A ∩ ran ⁡ B ⊆ dom ⁡ A ∩ F
5 fdm ⊢ A : C ⟶ D → dom ⁡ A = C
6 5 3ad2ant1 ⊢ A : C ⟶ D ∧ B : E ⟶ F ∧ C ∩ F = ∅ → dom ⁡ A = C
7 6 ineq1d ⊢ A : C ⟶ D ∧ B : E ⟶ F ∧ C ∩ F = ∅ → dom ⁡ A ∩ F = C ∩ F
8 simp3 ⊢ A : C ⟶ D ∧ B : E ⟶ F ∧ C ∩ F = ∅ → C ∩ F = ∅
9 7 8 eqtrd ⊢ A : C ⟶ D ∧ B : E ⟶ F ∧ C ∩ F = ∅ → dom ⁡ A ∩ F = ∅
10 4 9 sseqtrd ⊢ A : C ⟶ D ∧ B : E ⟶ F ∧ C ∩ F = ∅ → dom ⁡ A ∩ ran ⁡ B ⊆ ∅
11 ss0 ⊢ dom ⁡ A ∩ ran ⁡ B ⊆ ∅ → dom ⁡ A ∩ ran ⁡ B = ∅
12 10 11 syl ⊢ A : C ⟶ D ∧ B : E ⟶ F ∧ C ∩ F = ∅ → dom ⁡ A ∩ ran ⁡ B = ∅
13 12 coemptyd ⊢ A : C ⟶ D ∧ B : E ⟶ F ∧ C ∩ F = ∅ → A ∘ B = ∅