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Proof step 23 from proof max-rem-notsame (previous step)

bag0 -- bag0.max = ∅ ∨ bag0.max = (bag0 -- bag0.max).max ∨ count(bag0.max, bag0) = 1 ⊦ (bag0 = ∅ ⊃ m;(m < bag0.max ⊃ bag0.max;m)) = (m = (bag0 = ∅ ⊃ m;(m < bag0.max ⊃ bag0.max;m)) ⊃ bag0;(bag0 -- (bag0 = ∅ ⊃ m;bag0.max)) ++ m).max, (m = (bag0 = ∅ ⊃ m;(m < bag0.max ⊃ bag0.max;m)) ⊃ bag0;(bag0 -- (bag0 = ∅ ⊃ m;bag0.max)) ++ m) = ∅, bag0 = ∅, count((m < bag0.max ⊃ bag0.max;m), bag0 ++ m) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
bag0 = ∅ ++ bag0.max ∨ bag0.max = (bag0 -- bag0.max).max ∨ count(bag0.max, bag0) = 1 ⊦ (m = (m < bag0.max ⊃ bag0.max;m) ⊃ bag0;(bag0 -- bag0.max) ++ m) = ∅, (m < bag0.max ⊃ bag0.max;m) = (m = (m < bag0.max ⊃ bag0.max;m) ⊃ bag0;(bag0 -- bag0.max) ++ m).max, bag0 = ∅, count((m < bag0.max ⊃ bag0.max;m), bag0 ++ m) = 1

Rule argument

Additional information

Used simplifier rules: bag-rem-empty: ⊦ bag ≠ ∅ → (bag -- n = ∅ ↔ bag = ∅ ++ n);