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Proof step 5 from proof two-loup-05 (previous step)

m + n = m0 * 2 + n0, m0 + 1 < n, m0 + 1 < m, m < n, n0 < 2, 2 + m0 < n ⊦ m = 0, n = 0

User applied rule insert spec-lemma

And obtained the following new goals

Rule argument

spec: nat-basic2 inst: name: pls seq: m < m0, n < n0 ⊦ m + n + 1 < m0 + n0 subst: [m, m0, n, n0] → [m0 + 1, n, m0 + 1, m] flags:

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%two-02: ⊦ m + m = 2 * m; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%l-06: ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; from specification nat-basic2%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1 Used terms: m0 + 1, n, m0 + 1, m