Proof for mathd_algebra_513

[aleph-prover-test]

Proven lemmas: 1/1

The goal is to prove that for real numbers a, b ∈ ℝ satisfying the two equations 3a + 2b = 5 and a + b = 2, it follows that a = 1 and b = 1. The proof is naturally decomposed into two subgoals because the conclusion is a conjunction: first show a = 1, and then show b = 1. The key mathematical idea is to solve this 2×2 linear system: use a + b = 2 to express one variable in terms of the other (e.g. b = 2 − a), substitute into 3a + 2b = 5 to get a = 1, and then plug back to get b = 1. In Lean, this decomposition is implemented with constructor, and each of the two resulting linear goals is discharged automatically using linarith with the hypotheses h₀ and h₁. Progress-wise, the entire proof is complete: both subgoals (a = 1 and b = 1) have been solved. No remaining steps are left; the strategy is straightforward because linarith is well-suited to linear equalities and can solve the system directly without manual substitution.