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Metamath Proof Explorer

Theorem bicomdd

Description: Commute two sides of a biconditional in a deduction. (Contributed by Rodolfo Medina, 19-Oct-2010) (Proof shortened by Andrew Salmon, 29-Jun-2011)

Ref Expression
Hypothesis bicomdd.1 ⊢ φ → ψ → χ ↔ θ
Assertion bicomdd ⊢ φ → ψ → θ ↔ χ

Proof

Step Hyp Ref Expression
1 bicomdd.1 ⊢ φ → ψ → χ ↔ θ
2 bicom ⊢ χ ↔ θ ↔ θ ↔ χ
3 1 2 imbitrdi ⊢ φ → ψ → θ ↔ χ