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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 apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: square-ls-05 seq: ⊦ m * (m + m0) < n * (n + m0) ↔ m < n subst: [m, m0, n] → [n, m0, m]

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;%ls: ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≠ n → (m < n ↔ ¬ n < m); (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ ¬ n < n; (s) Used terms: n, m0, m