Proof for mathd_algebra_513 - #192
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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.
The proof is naturally decomposed into two subgoals: show a = 1 and show b = 1, since the conclusion is a conjunction a = 1 ∧ b = 1. In Lean, this is handled by first splitting the conjunction, then solving each equality separately.
Progress is complete: 2 out of 2 sub-problems are solved. The current proof uses linarith directly on the two hypotheses, and that is enough to derive both a = 1 and b = 1.
Nothing remains to be proved for this theorem. An interesting aspect is that there are also clean manual strategies: for example, use a + b = 2 to write b = 2 - a, substitute into 3a + 2b = 5, and get a = 1, then b = 1. Another useful check is that the coefficient matrix has nonzero determinant, so the system has a unique solution, which matches the result.