Proof for mathd_algebra_513 - #199
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Proven lemmas: 1/1
The goal is to prove that the real numbers a and b must both equal 1, assuming the two equations 3a + 2b = 5 and a + b = 2. Mathematically, this is just solving a 2×2 linear system over ℝ and showing the unique solution is (a, b) = (1, 1).
The proof was decomposed into the two component goals coming from the conjunction: first prove a = 1, then prove b = 1. Both of these are handled directly from the two hypotheses by linear elimination.
Progress is complete: 1 out of 1 theorem statements are solved, and within the theorem both subgoals have been discharged. In Lean, the proof uses constructor to split the conjunction and then linarith to solve each equality from the given linear equations.
Nothing remains open at this point. The key strategy is that h₀ − 2h₁ immediately gives a = 1, and then substituting into a + b = 2 gives b = 1; linarith automates exactly this kind of linear-arithmetic reasoning, so the final proof is very short.