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Proof step 6 from proof max-rem-notsame (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
(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, bag0.max = (bag0 -- bag0.max).max ∧ count(bag0.max, bag0) = 1 → bag0 -- bag0.max = ∅ ⊦ (m = (bag0 = ∅ ⊃ m;(m < bag0.max ⊃ bag0.max;m)) ⊃ bag0;(bag0 -- (bag0 = ∅ ⊃ m;bag0.max)) ++ m) = ∅

Rule argument

Additional information

Used simplifier rules: max-rec: ⊦ (bag ++ n).max = (bag = ∅ ⊃ n;(bag.max > n ⊃ bag.max;n));%bag-rem-empty: ⊦ bag ≠ ∅ → (bag -- n = ∅ ↔ bag = ∅ ++ n);%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ 1 = m + 1 ↔ m = 0; (s)% ⊦ count(n, bag0 ++ n) = count(n, bag0) + 1; (s)% ⊦ (bag0 ++ n) -- n0 = (n = n0 ⊃ bag0;(bag0 -- n0) ++ n); (s)% ⊦ m > n ↔ n < m; (s)