| Step | Hyp | Ref | Expression | 
						
							| 1 |  | ramval.c | ⊢ 𝐶  =  ( 𝑎  ∈  V ,  𝑖  ∈  ℕ0  ↦  { 𝑏  ∈  𝒫  𝑎  ∣  ( ♯ ‘ 𝑏 )  =  𝑖 } ) | 
						
							| 2 |  | 0nn0 | ⊢ 0  ∈  ℕ0 | 
						
							| 3 | 1 | hashbcval | ⊢ ( ( 𝐴  ∈  𝑉  ∧  0  ∈  ℕ0 )  →  ( 𝐴 𝐶 0 )  =  { 𝑥  ∈  𝒫  𝐴  ∣  ( ♯ ‘ 𝑥 )  =  0 } ) | 
						
							| 4 | 2 3 | mpan2 | ⊢ ( 𝐴  ∈  𝑉  →  ( 𝐴 𝐶 0 )  =  { 𝑥  ∈  𝒫  𝐴  ∣  ( ♯ ‘ 𝑥 )  =  0 } ) | 
						
							| 5 |  | hasheq0 | ⊢ ( 𝑥  ∈  V  →  ( ( ♯ ‘ 𝑥 )  =  0  ↔  𝑥  =  ∅ ) ) | 
						
							| 6 | 5 | elv | ⊢ ( ( ♯ ‘ 𝑥 )  =  0  ↔  𝑥  =  ∅ ) | 
						
							| 7 | 6 | anbi2i | ⊢ ( ( 𝑥  ∈  𝒫  𝐴  ∧  ( ♯ ‘ 𝑥 )  =  0 )  ↔  ( 𝑥  ∈  𝒫  𝐴  ∧  𝑥  =  ∅ ) ) | 
						
							| 8 |  | id | ⊢ ( 𝑥  =  ∅  →  𝑥  =  ∅ ) | 
						
							| 9 |  | 0elpw | ⊢ ∅  ∈  𝒫  𝐴 | 
						
							| 10 | 8 9 | eqeltrdi | ⊢ ( 𝑥  =  ∅  →  𝑥  ∈  𝒫  𝐴 ) | 
						
							| 11 | 10 | pm4.71ri | ⊢ ( 𝑥  =  ∅  ↔  ( 𝑥  ∈  𝒫  𝐴  ∧  𝑥  =  ∅ ) ) | 
						
							| 12 | 7 11 | bitr4i | ⊢ ( ( 𝑥  ∈  𝒫  𝐴  ∧  ( ♯ ‘ 𝑥 )  =  0 )  ↔  𝑥  =  ∅ ) | 
						
							| 13 | 12 | abbii | ⊢ { 𝑥  ∣  ( 𝑥  ∈  𝒫  𝐴  ∧  ( ♯ ‘ 𝑥 )  =  0 ) }  =  { 𝑥  ∣  𝑥  =  ∅ } | 
						
							| 14 |  | df-rab | ⊢ { 𝑥  ∈  𝒫  𝐴  ∣  ( ♯ ‘ 𝑥 )  =  0 }  =  { 𝑥  ∣  ( 𝑥  ∈  𝒫  𝐴  ∧  ( ♯ ‘ 𝑥 )  =  0 ) } | 
						
							| 15 |  | df-sn | ⊢ { ∅ }  =  { 𝑥  ∣  𝑥  =  ∅ } | 
						
							| 16 | 13 14 15 | 3eqtr4i | ⊢ { 𝑥  ∈  𝒫  𝐴  ∣  ( ♯ ‘ 𝑥 )  =  0 }  =  { ∅ } | 
						
							| 17 | 4 16 | eqtrdi | ⊢ ( 𝐴  ∈  𝑉  →  ( 𝐴 𝐶 0 )  =  { ∅ } ) |