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Proof step 6 from proof square-ls (previous step)

m0 * m0 + m0 < (n - 1) * (n - 1) + (n - 1) ↔ m0 + 1 < n ⊦ n = 0, 3 * m0 + m0 * m0 + 2 < n * n + n ↔ m0 + 1 < n

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
n = m + 1, m0 * m0 + m0 < m * m + m ↔ m0 + 1 < n ⊦ n = 0, 3 * m0 + m0 * m0 + 2 < n * n + n ↔ m0 + 1 < n

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m