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Proof step 15 from proof bag-rem-notmax (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 16:
bag0.max ≤ m0, ∀ m. (bag0 -- m).max = bag0.max ∨ m = bag0.max ⊦ (bag0 = ∅ ++ m ⊃ m0;(m0 < (bag0 -- m).max ⊃ (bag0 -- m).max;m0)) = m0, bag0 = ∅, bag0 = ∅ ++ m, m = m0

Rule argument

fma: bag0 = ∅ ++ m

Additional information

Used simplifier rules: ⊦ bag0 ++ n ≠ ∅; (s) Used formulas: bag0 = ∅ ++ m