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 the system 3a + 2b = 5 and a + b = 2, then necessarily a = 1 and b = 1. The proof is naturally decomposed into two sub-goals because the conclusion is a conjunction: first prove a = 1, then prove b = 1.

At this point, both sub-goals are solved, so progress is 2 out of 2 complete. Mathematically, the key idea is simple elimination: from a + b = 2, write b = 2 - a, substitute into 3a + 2b = 5, and simplify to get a = 1. Then substitute back into a + b = 2 to conclude b = 1.

In Lean, this is handled very efficiently by linarith, which can solve each equality directly from the two linear hypotheses. So there is no remaining work on this theorem: the proof is complete. The main interesting feature is that this is a straightforward linear system with a unique solution, and Lean’s linear arithmetic automation captures exactly that reasoning.