Metamath Proof Explorer
		
		
		
		Description:  A lemma for eliminating inequality, in inference form.  (Contributed by Giovanni Mascellani, 31-May-2019)
		
			
				
					|  |  | Ref | Expression | 
					
						|  | Hypotheses | sbceq1ddi.1 | ⊢ ( 𝜑  →  𝐴  =  𝐵 ) | 
					
						|  |  | sbceq1ddi.2 | ⊢ ( 𝜓  →  𝜃 ) | 
					
						|  |  | sbceq1ddi.3 | ⊢ ( [ 𝐴  /  𝑥 ] 𝜒  ↔  𝜃 ) | 
					
						|  |  | sbceq1ddi.4 | ⊢ ( [ 𝐵  /  𝑥 ] 𝜒  ↔  𝜂 ) | 
				
					|  | Assertion | sbceq1ddi | ⊢  ( ( 𝜑  ∧  𝜓 )  →  𝜂 ) | 
			
		
		
			
				Proof
				
					
						| Step | Hyp | Ref | Expression | 
						
							| 1 |  | sbceq1ddi.1 | ⊢ ( 𝜑  →  𝐴  =  𝐵 ) | 
						
							| 2 |  | sbceq1ddi.2 | ⊢ ( 𝜓  →  𝜃 ) | 
						
							| 3 |  | sbceq1ddi.3 | ⊢ ( [ 𝐴  /  𝑥 ] 𝜒  ↔  𝜃 ) | 
						
							| 4 |  | sbceq1ddi.4 | ⊢ ( [ 𝐵  /  𝑥 ] 𝜒  ↔  𝜂 ) | 
						
							| 5 | 1 | adantr | ⊢ ( ( 𝜑  ∧  𝜓 )  →  𝐴  =  𝐵 ) | 
						
							| 6 | 2 3 | sylibr | ⊢ ( 𝜓  →  [ 𝐴  /  𝑥 ] 𝜒 ) | 
						
							| 7 | 6 | adantl | ⊢ ( ( 𝜑  ∧  𝜓 )  →  [ 𝐴  /  𝑥 ] 𝜒 ) | 
						
							| 8 | 5 7 | sbceq1dd | ⊢ ( ( 𝜑  ∧  𝜓 )  →  [ 𝐵  /  𝑥 ] 𝜒 ) | 
						
							| 9 | 8 4 | sylib | ⊢ ( ( 𝜑  ∧  𝜓 )  →  𝜂 ) |