Metamath Proof Explorer


Theorem rncoeq

Description: Range of a composition. (Contributed by NM, 19-Mar-1998)

Ref Expression
Assertion rncoeq ⊢ dom ⁡ A = ran ⁡ B → ran ⁡ A ∘ B = ran ⁡ A

Proof

Step Hyp Ref Expression
1 dmcoeq ⊢ dom ⁡ B -1 = ran ⁡ A -1 → dom ⁡ B -1 ∘ A -1 = dom ⁡ A -1
2 eqcom ⊢ dom ⁡ A = ran ⁡ B ↔ ran ⁡ B = dom ⁡ A
3 df-rn ⊢ ran ⁡ B = dom ⁡ B -1
4 dfdm4 ⊢ dom ⁡ A = ran ⁡ A -1
5 3 4 eqeq12i ⊢ ran ⁡ B = dom ⁡ A ↔ dom ⁡ B -1 = ran ⁡ A -1
6 2 5 bitri ⊢ dom ⁡ A = ran ⁡ B ↔ dom ⁡ B -1 = ran ⁡ A -1
7 df-rn ⊢ ran ⁡ A ∘ B = dom ⁡ A ∘ B -1
8 cnvco ⊢ A ∘ B -1 = B -1 ∘ A -1
9 8 dmeqi ⊢ dom ⁡ A ∘ B -1 = dom ⁡ B -1 ∘ A -1
10 7 9 eqtri ⊢ ran ⁡ A ∘ B = dom ⁡ B -1 ∘ A -1
11 df-rn ⊢ ran ⁡ A = dom ⁡ A -1
12 10 11 eqeq12i ⊢ ran ⁡ A ∘ B = ran ⁡ A ↔ dom ⁡ B -1 ∘ A -1 = dom ⁡ A -1
13 1 6 12 3imtr4i ⊢ dom ⁡ A = ran ⁡ B → ran ⁡ A ∘ B = ran ⁡ A