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Proof step 3 from proof max-rem-notsame (previous step)

⊦ bag -- n = ∅, bag.max = (bag -- n).max ↔ n ≠ bag.max ∨ count(bag.max, bag) ≠ 1

User applied rule case distinction

And obtained the following new goals

Proof step 4:
bag.max = (bag -- n).max, count(bag.max, bag) = 1, n = bag.max ⊦ bag -- n = ∅
Proof step 19:
n ≠ bag.max ∨ count(bag.max, bag) ≠ 1 ⊦ bag -- n = ∅, bag.max = (bag -- n).max

Rule argument

pos: right 2

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