Metamath Proof Explorer


Theorem elimdelov

Description: Eliminate a hypothesis which is a predicate expressing membership in the result of an operator (deduction version). (Contributed by Paul Chapman, 25-Mar-2008)

Ref Expression
Hypotheses elimdelov.1 ( 𝜑𝐶 ∈ ( 𝐴 𝐹 𝐵 ) )
elimdelov.2 𝑍 ∈ ( 𝑋 𝐹 𝑌 )
Assertion elimdelov if ( 𝜑 , 𝐶 , 𝑍 ) ∈ ( if ( 𝜑 , 𝐴 , 𝑋 ) 𝐹 if ( 𝜑 , 𝐵 , 𝑌 ) )

Proof

Step Hyp Ref Expression
1 elimdelov.1 ( 𝜑𝐶 ∈ ( 𝐴 𝐹 𝐵 ) )
2 elimdelov.2 𝑍 ∈ ( 𝑋 𝐹 𝑌 )
3 iftrue ( 𝜑 → if ( 𝜑 , 𝐶 , 𝑍 ) = 𝐶 )
4 iftrue ( 𝜑 → if ( 𝜑 , 𝐴 , 𝑋 ) = 𝐴 )
5 iftrue ( 𝜑 → if ( 𝜑 , 𝐵 , 𝑌 ) = 𝐵 )
6 4 5 oveq12d ( 𝜑 → ( if ( 𝜑 , 𝐴 , 𝑋 ) 𝐹 if ( 𝜑 , 𝐵 , 𝑌 ) ) = ( 𝐴 𝐹 𝐵 ) )
7 1 3 6 3eltr4d ( 𝜑 → if ( 𝜑 , 𝐶 , 𝑍 ) ∈ ( if ( 𝜑 , 𝐴 , 𝑋 ) 𝐹 if ( 𝜑 , 𝐵 , 𝑌 ) ) )
8 iffalse ( ¬ 𝜑 → if ( 𝜑 , 𝐶 , 𝑍 ) = 𝑍 )
9 8 2 eqeltrdi ( ¬ 𝜑 → if ( 𝜑 , 𝐶 , 𝑍 ) ∈ ( 𝑋 𝐹 𝑌 ) )
10 iffalse ( ¬ 𝜑 → if ( 𝜑 , 𝐴 , 𝑋 ) = 𝑋 )
11 iffalse ( ¬ 𝜑 → if ( 𝜑 , 𝐵 , 𝑌 ) = 𝑌 )
12 10 11 oveq12d ( ¬ 𝜑 → ( if ( 𝜑 , 𝐴 , 𝑋 ) 𝐹 if ( 𝜑 , 𝐵 , 𝑌 ) ) = ( 𝑋 𝐹 𝑌 ) )
13 9 12 eleqtrrd ( ¬ 𝜑 → if ( 𝜑 , 𝐶 , 𝑍 ) ∈ ( if ( 𝜑 , 𝐴 , 𝑋 ) 𝐹 if ( 𝜑 , 𝐵 , 𝑌 ) ) )
14 7 13 pm2.61i if ( 𝜑 , 𝐶 , 𝑍 ) ∈ ( if ( 𝜑 , 𝐴 , 𝑋 ) 𝐹 if ( 𝜑 , 𝐵 , 𝑌 ) )