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Proof step 4 from proof two-loup-02 (previous step)

m + n = m0 * 2 + n0 ∧ n0 < 2 ∧ 2 ≠ 0, m0 - 1 < n, m < n, m0 - 1 < m, m + 1 < n, (m0 - 1) + 1 < n ⊦ m + 1 = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m + n = m0 * 2 + n0, m0 - 1 < n, m + 1 < n, m0 - 1 < m, m < n, n0 < 2, (m0 - 1) + 1 < n ⊦ m + 1 = n

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1; from specification nat-basic2%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%trans: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward) from specification nat-basic1