Proof for mathd_algebra_513

[aleph-prover-test]

Proven lemmas: 1/1

The goal is to prove that if a, b ∈ ℝ satisfy the linear system 3·a + 2·b = 5 and a + b = 2, then necessarily a = 1 and b = 1. The proof is decomposed by splitting the conjunction a = 1 ∧ b = 1 into two separate subgoals: first show a = 1, then show b = 1. For each subgoal, the idea is that the desired equality is a linear consequence of the two given equations, so Lean’s linear arithmetic tactic linarith can derive it directly from h₀ and h₁. In the current blueprint, both subgoals are already solved: after constructor creates the two goals, linarith [h₀, h₁] closes each one. Nothing remains to be proven; the proof is complete as written. An alternative strategy (also noted) would be to solve by substitution (derive b = 2 − a from a + b = 2, plug into 3a + 2b = 5 to get a = 1, then back-substitute for b), but the direct linarith approach is simpler and fully automated here.