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Proof step 5 from proof two-loup-02 (previous step)

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

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%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