| Step | Hyp | Ref | Expression | 
						
							| 1 |  | inass | ⊢ ( ( 𝐴  ∩  𝐵 )  ∩  ran  𝐹 )  =  ( 𝐴  ∩  ( 𝐵  ∩  ran  𝐹 ) ) | 
						
							| 2 |  | sseqin2 | ⊢ ( ran  𝐹  ⊆  𝐵  ↔  ( 𝐵  ∩  ran  𝐹 )  =  ran  𝐹 ) | 
						
							| 3 | 2 | biimpi | ⊢ ( ran  𝐹  ⊆  𝐵  →  ( 𝐵  ∩  ran  𝐹 )  =  ran  𝐹 ) | 
						
							| 4 | 3 | adantl | ⊢ ( ( Fun  𝐹  ∧  ran  𝐹  ⊆  𝐵 )  →  ( 𝐵  ∩  ran  𝐹 )  =  ran  𝐹 ) | 
						
							| 5 | 4 | ineq2d | ⊢ ( ( Fun  𝐹  ∧  ran  𝐹  ⊆  𝐵 )  →  ( 𝐴  ∩  ( 𝐵  ∩  ran  𝐹 ) )  =  ( 𝐴  ∩  ran  𝐹 ) ) | 
						
							| 6 | 1 5 | eqtrid | ⊢ ( ( Fun  𝐹  ∧  ran  𝐹  ⊆  𝐵 )  →  ( ( 𝐴  ∩  𝐵 )  ∩  ran  𝐹 )  =  ( 𝐴  ∩  ran  𝐹 ) ) | 
						
							| 7 | 6 | imaeq2d | ⊢ ( ( Fun  𝐹  ∧  ran  𝐹  ⊆  𝐵 )  →  ( ◡ 𝐹  “  ( ( 𝐴  ∩  𝐵 )  ∩  ran  𝐹 ) )  =  ( ◡ 𝐹  “  ( 𝐴  ∩  ran  𝐹 ) ) ) | 
						
							| 8 |  | fimacnvinrn | ⊢ ( Fun  𝐹  →  ( ◡ 𝐹  “  ( 𝐴  ∩  𝐵 ) )  =  ( ◡ 𝐹  “  ( ( 𝐴  ∩  𝐵 )  ∩  ran  𝐹 ) ) ) | 
						
							| 9 | 8 | adantr | ⊢ ( ( Fun  𝐹  ∧  ran  𝐹  ⊆  𝐵 )  →  ( ◡ 𝐹  “  ( 𝐴  ∩  𝐵 ) )  =  ( ◡ 𝐹  “  ( ( 𝐴  ∩  𝐵 )  ∩  ran  𝐹 ) ) ) | 
						
							| 10 |  | fimacnvinrn | ⊢ ( Fun  𝐹  →  ( ◡ 𝐹  “  𝐴 )  =  ( ◡ 𝐹  “  ( 𝐴  ∩  ran  𝐹 ) ) ) | 
						
							| 11 | 10 | adantr | ⊢ ( ( Fun  𝐹  ∧  ran  𝐹  ⊆  𝐵 )  →  ( ◡ 𝐹  “  𝐴 )  =  ( ◡ 𝐹  “  ( 𝐴  ∩  ran  𝐹 ) ) ) | 
						
							| 12 | 7 9 11 | 3eqtr4rd | ⊢ ( ( Fun  𝐹  ∧  ran  𝐹  ⊆  𝐵 )  →  ( ◡ 𝐹  “  𝐴 )  =  ( ◡ 𝐹  “  ( 𝐴  ∩  𝐵 ) ) ) |