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Proof step 4 from proof min-first (previous step)

n0 ≤ n2, n0 ≤ n1 ⊦ min(n0, min(n1, n2)) = n0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ min(n, m) = (n < m ⊃ n;m)
Proof step 6:
n0 ≤ n1, n0 ≤ n2 ⊦ (n0 < min(n1, n2) ⊃ n0;min(n1, n2)) = n0

Rule argument

spec: min-max name: min seq: ⊦ min(n, m) = (n < m ⊃ n;m) subst: [m, n] → [min(n1, n2), n0] flags: <all paths>

Additional information

Used terms: min(n1, n2), n0