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Proof step 11 from proof bag-rem-notmax (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 12:
m0 < bag0.max, ∀ m. (bag0 -- m).max = bag0.max ∨ m = bag0.max ⊦ (bag0 = ∅ ++ m ⊃ m0;(m0 < (bag0 -- m).max ⊃ (bag0 -- m).max;m0)) = bag0.max, bag0 = ∅, bag0 = ∅ ++ m, m = bag0.max, m0 = m

Rule argument

fma: bag0 = ∅ ++ m

Additional information

Used simplifier rules: max-one: ⊦ (∅ ++ n).max = n; Used formulas: bag0 = ∅ ++ m