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Proof step 35 from proof max-rem-notsame (previous step)

(bag = ∅ ⊃ m0;(m0 < bag.max ⊃ bag.max;m0)) < m, ∀ m. bag.max < m → bag = ∅ ∨ count(m, bag) = 0 ⊦ count(m, bag ++ m0) = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 36:
bag = ∅, m0 < m, ∀ m. bag.max < m → bag = ∅ ∨ count(m, bag) = 0 ⊦ count(m, bag ++ m0) = 0
Proof step 39:
(m0 < bag.max ⊃ bag.max;m0) < m, ∀ m. bag.max < m → bag = ∅ ∨ count(m, bag) = 0 ⊦ bag = ∅, count(m, bag ++ m0) = 0

Rule argument

pos: left 1

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