Metamath Proof Explorer


Theorem ifor

Description: Rewrite a disjunction in a conditional as two nested conditionals. (Contributed by Mario Carneiro, 28-Jul-2014)

Ref Expression
Assertion ifor ⊢ if φ ∨ ψ A B = if φ A if ψ A B

Proof

Step Hyp Ref Expression
1 iftrue ⊢ φ ∨ ψ → if φ ∨ ψ A B = A
2 1 orcs ⊢ φ → if φ ∨ ψ A B = A
3 iftrue ⊢ φ → if φ A if ψ A B = A
4 2 3 eqtr4d ⊢ φ → if φ ∨ ψ A B = if φ A if ψ A B
5 iffalse ⊢ ¬ φ → if φ A if ψ A B = if ψ A B
6 biorf ⊢ ¬ φ → ψ ↔ φ ∨ ψ
7 6 ifbid ⊢ ¬ φ → if ψ A B = if φ ∨ ψ A B
8 5 7 eqtr2d ⊢ ¬ φ → if φ ∨ ψ A B = if φ A if ψ A B
9 4 8 pm2.61i ⊢ if φ ∨ ψ A B = if φ A if ψ A B