Metamath Proof Explorer
		
		
		
		Description:  Equality deduction for nested function and operation value.
       (Contributed by AV, 23-Jul-2022)
		
			
				
					|  |  | Ref | Expression | 
					
						|  | Hypothesis | fvoveq1d.1 | ⊢ ( 𝜑  →  𝐴  =  𝐵 ) | 
				
					|  | Assertion | fvoveq1d | ⊢  ( 𝜑  →  ( 𝐹 ‘ ( 𝐴 𝑂 𝐶 ) )  =  ( 𝐹 ‘ ( 𝐵 𝑂 𝐶 ) ) ) | 
			
		
		
			
				Proof
				
					
						| Step | Hyp | Ref | Expression | 
						
							| 1 |  | fvoveq1d.1 | ⊢ ( 𝜑  →  𝐴  =  𝐵 ) | 
						
							| 2 | 1 | oveq1d | ⊢ ( 𝜑  →  ( 𝐴 𝑂 𝐶 )  =  ( 𝐵 𝑂 𝐶 ) ) | 
						
							| 3 | 2 | fveq2d | ⊢ ( 𝜑  →  ( 𝐹 ‘ ( 𝐴 𝑂 𝐶 ) )  =  ( 𝐹 ‘ ( 𝐵 𝑂 𝐶 ) ) ) |