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Proof step 17 from proof bag-rem-notmax (previous step)

bag0.max ≤ m0, ∀ m. (bag0 -- m).max = bag0.max ∨ m = bag0.max ⊦ (m0 < (bag0 -- m).max ⊃ (bag0 -- m).max;m0) = m0, bag0 = ∅ ++ m, bag0 = ∅, m = m0

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 18:
m0 < (bag0 -- m).max, bag0.max ≤ m0, ∀ m. (bag0 -- m).max = bag0.max ∨ m = bag0.max ⊦ (m0 < (bag0 -- m).max ⊃ (bag0 -- m).max;m0) = m0, bag0 = ∅, bag0 = ∅ ++ m, m = m0

Rule argument

fma: m0 < (bag0 -- m).max

Additional information

Used formulas: m0 < (bag0 -- m).max