Proof for mathd_algebra_513 - #206
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Proven lemmas: 1/1
The goal is to prove that if a, b ∈ ℝ satisfy the system 3a + 2b = 5 and a + b = 2, then necessarily a = 1 and b = 1.
Mathematically, this was split into the two natural subgoals coming from the conjunction: first prove a = 1, then prove b = 1. The key idea is simple linear elimination: double the equation a + b = 2 to get 2a + 2b = 4, subtract from 3a + 2b = 5, and obtain a = 1; then substitute back into a + b = 2 to get b = 1.
Progress is complete: 1 out of 1 theorem goals has been solved, and both component facts a = 1 and b = 1 are established. There is nothing remaining to prove.
An interesting point is that Lean can discharge both parts very efficiently using linarith, since the statement is just a small linear system over ℝ. So although the proof can be written out by hand with elimination, the formal proof is essentially immediate once the conjunction is split.