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Proof step 1 from proof min-distr (conclusion)

⊦ min(m + 1, n + 1) = min(m, n) + 1

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 2:
⊦ m + 1 < n + 1, min(m + 1, n + 1) = min(m, n) + 1

Rule argument

fma: m + 1 < n + 1

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);%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2 Used formulas: m + 1 < n + 1