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Proof step 15 from proof min-a (previous step)

min(m, n) < n, min(m, n) < m, n0 ≤ n, n0 ≤ m, min(m, n) < n0 ⊦ n0 = 0, n0 = min(m, n), n = 0, m = 0

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: m < n

Additional information

Used simplifier rules: Min-left: ⊦ m < n → min(m, n) = m;%Min-right: ⊦ ¬ m < n → min(m, n) = n;%ref-1: ⊦ ¬ n < n; from specification nat-basic1 Used formulas: m < n