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Proof step 7 from proof two-loup-04 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
2 + m < 0 - 1, m < 0 - 1, m + 1 < 0 - 1, m + n < 2, m + 1 < n, m < n, n < 0 - 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% ⊦ 0 + m = m; (s)