Proof for mathd_algebra_513

[aleph-prover-test]

Proven lemmas: 3/3

The goal is to prove that for real numbers a, b ∈ ℝ satisfying the linear system 3a + 2b = 5 and a + b = 2, it follows that a = 1 ∧ b = 1.
The proof is decomposed into two sub-results: one lemma proves a = 1 from the two equations, and a second lemma proves b = 1 from the same assumptions.
Both sub-lemmas are already solved: each is discharged by Lean’s linear arithmetic tactic linarith, which can solve small linear systems automatically.
With these in hand, the main theorem is proved by combining the two conclusions using a conjunction constructor, yielding ⟨a = 1, b = 1⟩.
So progress is complete: 2 sub-problems solved out of 2, and the final statement is also finished.
An interesting aspect is that the whole argument can be done either “explicitly” by eliminating variables (e.g., doubling a + b = 2 and subtracting from 3a + 2b = 5) or fully automatically via linarith, and the current proof uses the automated route.