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Proof step 11 from proof min-a (previous step)

n < min(m, n), n0 ≤ min(m, n), n ≤ m, n < n0 ⊦ n = n0, n0 = 0, min(m, n) = 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%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat Used formulas: m < n