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Proof step 5 from proof two-loup-04 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m + n = m0 * 2 + n0, n < m0 - 1, m < n, m + 1 < n, n0 < 2, m + 1 < m0 - 1, m < m0 - 1, 2 + m < m0 - 1 ⊦ m + 1 = n

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%trans-01: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic