Proof for mathd_algebra_513

[aleph-prover-dev]

Proven lemmas: 1/1

The goal is to prove that if real numbers a and b satisfy the linear system 3·a + 2·b = 5 and a + b = 2, then necessarily a = 1 and b = 1 (i.e., the unique solution is (1,1)). The proof is decomposed by splitting the final conjunction into two subgoals: first show a = 1, and then show b = 1. Lean does this with constructor, producing exactly 2 sub-problems from the original goal. Both subgoals are then solved automatically using linarith with the hypotheses h₀ and h₁, which eliminates variables in linear equalities/inequalities and derives the needed equalities. So progress is complete: 2 out of 2 sub-problems are solved, and nothing remains. An interesting aspect is that this is a standard “solve a 2×2 linear system” situation; automation (linarith) is sufficient, but a manual fallback strategy would be to solve for b from a + b = 2, substitute into 3·a + 2·b = 5, and then back-substitute. The system’s coefficient matrix has nonzero determinant, so the solution is unique, which matches the derived result.