Proof for mathd_algebra_513

[aleph-prover-test]

Proven lemmas: 1/1

The goal is to prove that if real numbers 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.

The proof was naturally decomposed into the two target conclusions a = 1 and b = 1, since the theorem asks for a conjunction. From there, one can eliminate one variable using a + b = 2 and substitute into 3a + 2b = 5, or let Lean’s linear arithmetic tactic handle both equations directly.

Progress is complete: 1 out of 1 theorem goals has been solved. In particular, Lean can finish the proof by splitting the conjunction and using linarith on the two hypotheses to derive each equality.

There is nothing substantial remaining to prove. An explicit human-style route is: from a + b = 2, write a = 2 - b; substitute into 3a + 2b = 5 to get 3(2 - b) + 2b = 5, hence 6 - b = 5, so b = 1, and then a = 1.

The main interesting point is that this is a very clean example where a general linear-arithmetic strategy in Lean matches the standard algebraic solution exactly. No obstacles or inconsistencies appeared; the theorem is correct and has a short proof.