| Step | Hyp | Ref | Expression | 
						
							| 1 |  | srgz.b | ⊢ 𝐵  =  ( Base ‘ 𝑅 ) | 
						
							| 2 |  | srgz.t | ⊢  ·   =  ( .r ‘ 𝑅 ) | 
						
							| 3 |  | srgz.z | ⊢  0   =  ( 0g ‘ 𝑅 ) | 
						
							| 4 |  | srgisid.1 | ⊢ ( 𝜑  →  𝑅  ∈  SRing ) | 
						
							| 5 |  | srgisid.2 | ⊢ ( 𝜑  →  𝑍  ∈  𝐵 ) | 
						
							| 6 |  | srgisid.3 | ⊢ ( ( 𝜑  ∧  𝑥  ∈  𝐵 )  →  ( 𝑍  ·  𝑥 )  =  𝑍 ) | 
						
							| 7 | 6 | ralrimiva | ⊢ ( 𝜑  →  ∀ 𝑥  ∈  𝐵 ( 𝑍  ·  𝑥 )  =  𝑍 ) | 
						
							| 8 | 1 3 | srg0cl | ⊢ ( 𝑅  ∈  SRing  →   0   ∈  𝐵 ) | 
						
							| 9 |  | oveq2 | ⊢ ( 𝑥  =   0   →  ( 𝑍  ·  𝑥 )  =  ( 𝑍  ·   0  ) ) | 
						
							| 10 | 9 | eqeq1d | ⊢ ( 𝑥  =   0   →  ( ( 𝑍  ·  𝑥 )  =  𝑍  ↔  ( 𝑍  ·   0  )  =  𝑍 ) ) | 
						
							| 11 | 10 | rspcv | ⊢ (  0   ∈  𝐵  →  ( ∀ 𝑥  ∈  𝐵 ( 𝑍  ·  𝑥 )  =  𝑍  →  ( 𝑍  ·   0  )  =  𝑍 ) ) | 
						
							| 12 | 4 8 11 | 3syl | ⊢ ( 𝜑  →  ( ∀ 𝑥  ∈  𝐵 ( 𝑍  ·  𝑥 )  =  𝑍  →  ( 𝑍  ·   0  )  =  𝑍 ) ) | 
						
							| 13 | 7 12 | mpd | ⊢ ( 𝜑  →  ( 𝑍  ·   0  )  =  𝑍 ) | 
						
							| 14 | 1 2 3 | srgrz | ⊢ ( ( 𝑅  ∈  SRing  ∧  𝑍  ∈  𝐵 )  →  ( 𝑍  ·   0  )  =   0  ) | 
						
							| 15 | 4 5 14 | syl2anc | ⊢ ( 𝜑  →  ( 𝑍  ·   0  )  =   0  ) | 
						
							| 16 | 13 15 | eqtr3d | ⊢ ( 𝜑  →  𝑍  =   0  ) |