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Proof step 2 from proof min-ls-both (previous step)

min(m, n) < m, min(m, n) < n ⊦ m = 0, 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-a: ⊦ min(m, min(n, n0)) = min(min(m, n), n0);%min-c: ⊦ min(m, n) = min(n, m);%Min-right: ⊦ ¬ m < n → min(m, n) = n;%min-a: ⊦ min(m, min(n, n0)) = min(min(m, n), n0);%min-c: ⊦ min(m, n) = min(n, m);%Min-left: ⊦ m < n → min(m, n) = m;%ref-1: ⊦ ¬ n < n; from specification nat-basic1 Used formulas: m < n