Proof for mathd_algebra_513, eq_four

[aleph-prover-test]

Proven lemmas: 4/4

The main goal is to prove the real-number system of equations result: for a, b ∈ ℝ, if 3a + 2b = 5 and a + b = 2, then a = 1 and b = 1 (i.e., a = 1 ∧ b = 1). This was decomposed into two simpler sub-problems: first prove a = 1 from the two linear equations, and then prove b = 1 from the same hypotheses. Both sub-lemmas are completed using Lean’s linear arithmetic tactic linarith, which can directly solve these simultaneous linear equations. The final theorem mathd_algebra_513 is then finished by combining the two sub-results with a conjunction constructor.

There is also a separate theorem eq_four: ∀ a b c d ∈ Nat, if a = b, a = d, and a = c, then c = b. This has already been proved by a simple equality chain: from a = c get c = a by symmetry, then compose with a = b to obtain c = b; the hypothesis a = d is unused.

Overall, all identified pieces are solved: 4 statements total (the two helper lemmas, the combined main theorem, and eq_four), and all 4 are complete. Nothing remains unfinished at this stage. The interesting strategy point is that linarith cleanly eliminates the need for manual algebra in the linear system, while eq_four demonstrates a minimal “symmetry + transitivity” equality proof.