Proof for mathd_algebra_513

[aleph-prover-test]

Proven lemmas: 1/1

The current goal is to prove the theorem: if a, b ∈ ℝ satisfy 3a + 2b = 5 and a + b = 2, then necessarily a = 1 and b = 1. Mathematically, this is just solving a 2×2 linear system and showing the unique solution is (1, 1).

The proof was not really split into separate helper lemmas; it was decomposed only into the two coordinates of the final conjunction, namely proving a = 1 and proving b = 1. Both of these subgoals have already been solved.

So the progress is complete: 1 theorem out of 1 has been proved, and both internal sub-steps are finished. There is no remaining mathematical work on this statement.

The main strategy is very simple: use the two linear equations together. One can subtract 2(a + b) = 4 from 3a + 2b = 5 to get a = 1, then substitute back into a + b = 2 to obtain b = 1; in Lean, linarith handles these linear deductions directly. There were no real obstacles here, since the system is linear and has a unique solution.