Metamath Proof Explorer


Theorem ifcld

Description: Membership (closure) of a conditional operator, deduction form. (Contributed by SO, 16-Jul-2018)

Ref Expression
Hypotheses ifcld.a ⊢ ( 𝜑 → 𝐴 ∈ 𝐶 )
ifcld.b ⊢ ( 𝜑 → 𝐵 ∈ 𝐶 )
Assertion ifcld ( 𝜑 → if ( 𝜓 , 𝐴 , 𝐵 ) ∈ 𝐶 )

Proof

Step Hyp Ref Expression
1 ifcld.a ⊢ ( 𝜑 → 𝐴 ∈ 𝐶 )
2 ifcld.b ⊢ ( 𝜑 → 𝐵 ∈ 𝐶 )
3 ifcl ⊢ ( ( 𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐶 ) → if ( 𝜓 , 𝐴 , 𝐵 ) ∈ 𝐶 )
4 1 2 3 syl2anc ⊢ ( 𝜑 → if ( 𝜓 , 𝐴 , 𝐵 ) ∈ 𝐶 )