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Proof step 5 from proof bag-rem-notmax (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
∀ m. (bag0 -- m).max = bag0.max ∨ m = bag0.max ⊦ (m0 = m ⊃ bag0;(bag0 -- m) ++ m0).max = (bag0 = ∅ ⊃ m0;(m0 < bag0.max ⊃ bag0.max;m0)), bag0 = ∅, m = (bag0 = ∅ ⊃ m0;(m0 < bag0.max ⊃ bag0.max;m0))

Rule argument

fma: bag0 = ∅

Additional information

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