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Metamath Proof Explorer

Theorem elimif

Description: Elimination of a conditional operator contained in a wff ps . (Contributed by NM, 15-Feb-2005) (Proof shortened by NM, 25-Apr-2019)

Ref Expression
Hypotheses elimif.1 ⊢ if φ A B = A → ψ ↔ χ
elimif.2 ⊢ if φ A B = B → ψ ↔ θ
Assertion elimif ⊢ ψ ↔ φ ∧ χ ∨ ¬ φ ∧ θ

Proof

Step Hyp Ref Expression
1 elimif.1 ⊢ if φ A B = A → ψ ↔ χ
2 elimif.2 ⊢ if φ A B = B → ψ ↔ θ
3 iftrue ⊢ φ → if φ A B = A
4 3 1 syl ⊢ φ → ψ ↔ χ
5 iffalse ⊢ ¬ φ → if φ A B = B
6 5 2 syl ⊢ ¬ φ → ψ ↔ θ
7 4 6 cases ⊢ ψ ↔ φ ∧ χ ∨ ¬ φ ∧ θ