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Proof step 22 from proof max-rem-notsame (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 23:
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

Rule argument

pos: right 3

Additional information

Used simplifier rules: ⊦ count(n, ∅ ++ n) = 1; (s)