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Proof step 2 from proof min-distr (previous step)

⊦ m + 1 < n + 1, min(m + 1, n + 1) = min(m, n) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n ≤ m ⊦ n = min(m, n)

Rule argument

Additional information

Used simplifier rules: 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);%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2