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Proof step 21 from proof max-rem-notsame (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
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

Rule argument

Additional information

Used simplifier rules: max-rec: ⊦ (bag ++ n).max = (bag = ∅ ⊃ n;(bag.max > n ⊃ bag.max;n));% ⊦ (bag0 ++ n) -- n0 = (n = n0 ⊃ bag0;(bag0 -- n0) ++ n); (s)% ⊦ m > n ↔ n < m; (s)