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Proof step 11 from proof anti (previous step)

n = n0 +1, m ≤ n, n ≤ m ⊦ m = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
n0 + 1 < m, n0 < m, m ≤ n0 + 1 ⊦ m = n0 + 1

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ m + 1 ≤ n ↔ m < n;% ⊦ m < n → n ≠ 0; (s)%succ-03: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%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)