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Proof step 6 from proof two-loup-04 (previous step)

m0 = m1 + 1, m + n = m0 * 2 + n0, n0 < 2, 2 + m < m0, m + 1 < n, m + 1 < m0, m < n, n + 1 < m0 ⊦ n = 0, m1 = 0, m + 1 = n, m0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m + n = (2 * m1 + 2) + n0, n < m1, m < n, m + 1 < n, n0 < 2, m + 1 < m1, m < m1, 2 + m < m1 ⊦ m + 1 = n, m1 = 0, n = 0

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1; from specification nat-basic2%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%l-08: ⊦ m + n < m0 → m + n < m0 + n0; from specification nat-basic2%l-num-02: n0 < n ⊦ m + (* n) < m0 + (* n0) ↔ m + (n - n0) < m0; from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%rsucc-02: ⊦ m * (n + 1) = m * n + m; from specification nat-mult