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Proof step 3 from proof min-0 (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 4:
min(n0, min(n1, n2)) = 0 ⊦ n0 = 0, n1 = 0, n2 = 0

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ min(0, n) = 0; (s)% ⊦ min(n, 0) = 0; (s)