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Proof step 6 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)), bag0 = ∅, m = (bag0 = ∅ ⊃ m0;(m0 < bag0.max ⊃ bag0.max;m0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ m. (bag0 -- m).max = bag0.max ∨ m = bag0.max ⊦ (m0 = m ⊃ bag0;(bag0 -- m) ++ m0).max = (m0 < bag0.max ⊃ bag0.max;m0), bag0 = ∅, m = (m0 < bag0.max ⊃ bag0.max;m0)

Proof step 6 from proof bag-rem-notmax (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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