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Proof step 3 from proof min-le-diff-third (previous step)

⊦ n2 ≤ n2 - min(n0, min(n1, n2)) ↔ n0 = 0 ∨ n1 = 0 ∨ n2 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ min(n, m) = (n < m ⊃ n;m)
Proof step 5:
⊦ n2 ≤ n2 - (n0 < min(n1, n2) ⊃ n0;min(n1, n2)) ↔ n0 = 0 ∨ n1 = 0 ∨ n2 = 0

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