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Proof step 5 from proof square-nls (previous step)

n * n + n = m0 * 2 + n0, n0 < 2, m0 < n ⊦ n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-basic1 inst: name: twols-01 seq: ⊦ m ≠ 1 → (m < 2 ↔ m = 0) subst: [m] → [n0] flags: rewrite

Additional information

Used simplifier rules: add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%twoeqf: ⊦ m + m0 = 2 * n ∧ n < m → m0 < n; (forward) from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%twoeqf-01: ⊦ m + m0 = 2 * n ∧ m < n → n < m0; (forward) from specification nat-mult%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%succ: ⊦ m +1 = m + 1; from specification nat-basic2%ls-08: ⊦ m * n + n0 < n ↔ m = 0 ∧ n0 < n; from specification nat-mult%square-even-01: ⊦ m * m + m ≠ 2 * n + 1; from specification nat-mult Used terms: n0