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Proof step 19 from proof max-a (previous step)

max(m, n) < n, max(m, n) < m, n0 < m, n0 ≤ n, n ≤ m, max(m, n) < n0, max(m, n) + 1 < m ⊦ m = 0, n0 = 0, m = n0, n = 0

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: m < n

Additional information

Used simplifier rules: Max-right: ⊦ ¬ m < n → max(m, n) = m;%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat Used formulas: m < n