| Step | Hyp | Ref | Expression | 
						
							| 1 |  | ancom | ⊢ ( ( 𝐴  ⊆  𝐵  ∧  𝐵  ⊆  𝐴 )  ↔  ( 𝐵  ⊆  𝐴  ∧  𝐴  ⊆  𝐵 ) ) | 
						
							| 2 |  | sscon34b | ⊢ ( ( 𝐵  ⊆  𝐶  ∧  𝐴  ⊆  𝐶 )  →  ( 𝐵  ⊆  𝐴  ↔  ( 𝐶  ∖  𝐴 )  ⊆  ( 𝐶  ∖  𝐵 ) ) ) | 
						
							| 3 | 2 | ancoms | ⊢ ( ( 𝐴  ⊆  𝐶  ∧  𝐵  ⊆  𝐶 )  →  ( 𝐵  ⊆  𝐴  ↔  ( 𝐶  ∖  𝐴 )  ⊆  ( 𝐶  ∖  𝐵 ) ) ) | 
						
							| 4 |  | sscon34b | ⊢ ( ( 𝐴  ⊆  𝐶  ∧  𝐵  ⊆  𝐶 )  →  ( 𝐴  ⊆  𝐵  ↔  ( 𝐶  ∖  𝐵 )  ⊆  ( 𝐶  ∖  𝐴 ) ) ) | 
						
							| 5 | 3 4 | anbi12d | ⊢ ( ( 𝐴  ⊆  𝐶  ∧  𝐵  ⊆  𝐶 )  →  ( ( 𝐵  ⊆  𝐴  ∧  𝐴  ⊆  𝐵 )  ↔  ( ( 𝐶  ∖  𝐴 )  ⊆  ( 𝐶  ∖  𝐵 )  ∧  ( 𝐶  ∖  𝐵 )  ⊆  ( 𝐶  ∖  𝐴 ) ) ) ) | 
						
							| 6 | 1 5 | bitrid | ⊢ ( ( 𝐴  ⊆  𝐶  ∧  𝐵  ⊆  𝐶 )  →  ( ( 𝐴  ⊆  𝐵  ∧  𝐵  ⊆  𝐴 )  ↔  ( ( 𝐶  ∖  𝐴 )  ⊆  ( 𝐶  ∖  𝐵 )  ∧  ( 𝐶  ∖  𝐵 )  ⊆  ( 𝐶  ∖  𝐴 ) ) ) ) | 
						
							| 7 |  | eqss | ⊢ ( 𝐴  =  𝐵  ↔  ( 𝐴  ⊆  𝐵  ∧  𝐵  ⊆  𝐴 ) ) | 
						
							| 8 |  | eqss | ⊢ ( ( 𝐶  ∖  𝐴 )  =  ( 𝐶  ∖  𝐵 )  ↔  ( ( 𝐶  ∖  𝐴 )  ⊆  ( 𝐶  ∖  𝐵 )  ∧  ( 𝐶  ∖  𝐵 )  ⊆  ( 𝐶  ∖  𝐴 ) ) ) | 
						
							| 9 | 6 7 8 | 3bitr4g | ⊢ ( ( 𝐴  ⊆  𝐶  ∧  𝐵  ⊆  𝐶 )  →  ( 𝐴  =  𝐵  ↔  ( 𝐶  ∖  𝐴 )  =  ( 𝐶  ∖  𝐵 ) ) ) |