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Proof step 7 from proof max-rem-notsame (previous step)

(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) = ∅

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 8:
(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) = ∅, bag0 = ∅

Rule argument

fma: bag0 = ∅

Additional information

Used simplifier rules: ⊦ count(n, ∅ ++ n) = 1; (s) Used formulas: bag0 = ∅