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Proof step 5 from proof square-eq-03 (previous step)

m * (m + m0) = n * (n + m0) ⊦ m < n, m = n

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: nat-mult inst: name: square-ls-05 seq: ⊦ m * (m + m0) < n * (n + m0) ↔ m < n subst: [m, m0, n] → [n, m0, m] flags: current, []

Additional information

Used simplifier rules: dist: ⊦ m0 * (m + n) = m0 * m + m0 * n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%neq-01: ⊦ m ≠ n → (¬ m < n ↔ n < m); from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1 Used terms: n, m0, m