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Proof step 40 from proof max-rem-notsame (previous step)

(m0 < bag.max ⊃ bag.max;m0) < m, ∀ m1. bag.max < m1 → count(m1, bag) = 0 ⊦ bag = ∅, count(m, bag ++ m0) = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 41:
bag.max < m, m0 < bag.max, ∀ m1. bag.max < m1 → count(m1, bag) = 0 ⊦ bag = ∅, count(m, bag ++ m0) = 0
Proof step 45:
m0 < m, ∀ m1. bag.max < m1 → count(m1, bag) = 0 ⊦ bag = ∅, count(m, bag ++ m0) = 0, m0 < bag.max

Rule argument

pos: left 1

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