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Proof step 3 from proof square-inject (previous step)

m * (m + 1) + 2 * m0 = n * (n + 1) + 2 * n0, m0 ≤ m, n0 ≤ n ⊦ m * (m + 1) ≤ n * (n + 1) + 2 * n0 ∧ n * (n + 1) + 2 * n0 < (m + 1) * (m + 2), m = n ∧ m0 = n0

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 1 flags: rotate, discard

Additional information

Used simplifier rules: rsucc-01: ⊦ (m + 1) * n = m * n + n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%dist: ⊦ m0 * (m + n) = m0 * m + m0 * n;%succ: ⊦ (* m0) * m + m = (m0 + 1) * m;%numeral-24: m1 ≠ 0 ⊦ m + n < (* m1) * m + n0 ↔ n < (m1 - 1) * m + n0;%coef-21: ⊦ m * n < m1 * n + n ↔ ¬ ¬(m < m1 + 1 ∧ n ≠ 0);%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%lep: ⊦ m ≤ m + n; from specification nat%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat