lean_workbook_plus_2370

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Disproved status carries no information about that problem. Do not import this node or use it as a dependency.

From the condition we can let $a=cosA$ , $b=cosB$ , $c=cosC$ where $A, B, C$ are the angles of a triangle. Using $\sqrt{\frac{1-cosA}{1+cosA}}=tan\frac{A}{2}$ $\sqrt{\frac{(1-a)(1-b)}{(1+a)(1+b)}}+\sqrt{\frac{(1-b)(1-c)}{(1+b)(1+c)}}+\sqrt{\frac{(1-c)(1-a)}{(1+c)(1+a)}}=1$ becomes $tan\frac{A}{2}tan\frac{B}{2}+tan\frac{B}{2}tan\frac{C}{2}+tan\frac{C}{2}tan\frac{A}{2}=1$ Which is well known identity.

Why this node was retired

The posted statement is

theorem lean_workbook_plus_2370 :
  ∀ a b c : ℝ, a > 0 ∧ b > 0 ∧ c > 0 ∧ a + b > c ∧ a + c > b ∧ b + c > a →
  Real.sqrt ((1 - a) * (1 - b) / ((1 + a) * (1 + b))) +
  Real.sqrt ((1 - b) * (1 - c) / ((1 + b) * (1 + c))) +
  Real.sqrt ((1 - c) * (1 - a) / ((1 + c) * (1 + a))) = 1   :=  by sorry

Define a=cosA,b=cosB,c=cosC with A,B,C>0 and A+B+C=π; do not replace angle cosines by triangle side lengths.

The recorded counterexample refutes the statement as encoded. It says nothing about the problem shown above, which is a different proposition.

Proposed corrected statement

Define a=cosA,b=cosB,c=cosC with A,B,C>0 and A+B+C=π; do not replace angle cosines by triangle side lengths.

Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.