Proof for mathd_algebra_513

[aleph-prover-test]

Proven lemmas: 1/1

The current theorem proves 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 naturally decomposes into two sub-goals: show a = 1 and show b = 1, since the target is a conjunction a = 1 ∧ b = 1.

Progress is complete: 2 out of 2 sub-problems are solved. The main strategy is straightforward linear algebra/arithmetic: from a + b = 2, rewrite b = 2 - a, substitute into 3a + 2b = 5, and solve to get a = 1; then plug back in to get b = 1. In Lean, this is handled very efficiently by linarith, which can solve both linear equalities directly from the hypotheses.

So there is nothing remaining to prove for this node. An interesting aspect is that although the informal solution is just substitution, the formal proof can be compressed into a one-line linear-arithmetic argument, reflecting that the system has a unique solution.