Proof for mathd_algebra_513

[aleph-prover]

Proven lemmas: 1/1

The goal is to prove that if real numbers a and b satisfy the system 3a + 2b = 5 and a + b = 2, then necessarily a = 1 and b = 1. In other words, this is a simple 2×2 linear system over ℝ, and the theorem states its unique solution.

The proof was decomposed into the two parts of the conjunction: first prove a = 1, then prove b = 1. Both subgoals can be handled directly from the two given equations using linear arithmetic.

Progress is complete: 2 out of 2 sub-problems are solved. Lean’s linarith tactic is enough for each part once the conjunction is split with constructor.

The key idea is that subtracting 2·(a + b = 2) from 3a + 2b = 5 gives a = 1, and then substituting into a + b = 2 gives b = 1. An interesting point is that the coefficient matrix has nonzero determinant, so the system has a unique solution, which matches the computed answer (1, 1). There do not seem to be any remaining obstacles or unfinished parts.