Proof for mathd_algebra_513 - #58
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Proven lemmas: 1/1
The goal is to prove that for real numbers a, b ∈ ℝ satisfying the linear system 3·a + 2·b = 5 and a + b = 2, the unique solution is a = 1 and b = 1 (so the conclusion is a = 1 ∧ b = 1).
The proof is decomposed into two main sub-goals: first derive ha : a = 1 from the two given equations, and then derive hb : b = 1 from the same equations, finally pairing them to get the conjunction.
Both sub-goals have been completed: Lean uses the linarith tactic twice, once to solve for a and once to solve for b, directly from h₀ and h₁.
With ha and hb established, the proof finishes by returning ⟨ha, hb⟩, which exactly matches the desired statement a = 1 ∧ b = 1.
So progress is complete: 2 sub-problems out of 2 solved, with nothing remaining.
An interesting aspect is that linarith automatically performs the needed linear algebra (equivalent to taking suitable linear combinations of the equations), avoiding manual substitution or arithmetic.