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 linear equations 3·a + 2·b = 5 and a + b = 2, it follows that a = 1 and b = 1 (i.e., a = 1 ∧ b = 1).
The proof is decomposed into two sub-goals because the conclusion is a conjunction: first show a = 1, then show b = 1.
So far, both sub-goals have been solved: Lean uses constructor to split the goal into the two parts, and then applies linarith [h₀, h₁] to each part to derive the required equalities from the given linear system.
That means progress is complete: 2 out of 2 sub-problems are finished, and there is nothing remaining to prove.
An interesting aspect is that linarith can solve each variable directly from the pair of equations without manual substitution; alternatively, one could solve by substituting b = 2 − a from a + b = 2 into the first equation and back-substituting, but linarith automates this linear-algebra step.
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