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Proof step 4 from proof two-loup-11 (previous step)

m + n = m0 * 2 + n0 ∧ n0 < 2 ∧ 2 ≠ 0, m < n, m0 ≤ m, m + 1 < n, m0 < n ⊦ m + 1 = n, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m + n = m0 * 2 + n0, n0 < 2, m < n, m0 ≤ m, m + 1 < n, m0 < n ⊦ m + 1 = n, n = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat