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Proof step 9 from proof max-rem-notsame (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 10:
(m < bag0.max ⊃ bag0.max;m) = (m = (m < bag0.max ⊃ bag0.max;m) ⊃ bag0;(bag0 -- bag0.max) ++ m).max, count((m < bag0.max ⊃ bag0.max;m), bag0 ++ m) = 1, m < bag0.max, bag0.max = (bag0 -- bag0.max).max ∧ count(bag0.max, bag0) = 1 → bag0 = ∅ ++ bag0.max ⊦ (m = (m < bag0.max ⊃ bag0.max;m) ⊃ bag0;(bag0 -- bag0.max) ++ m) = ∅, bag0 = ∅
Proof step 12:
(m < bag0.max ⊃ bag0.max;m) = (m = (m < bag0.max ⊃ bag0.max;m) ⊃ bag0;(bag0 -- bag0.max) ++ m).max, count((m < bag0.max ⊃ bag0.max;m), bag0 ++ m) = 1, bag0.max = (bag0 -- bag0.max).max ∧ count(bag0.max, bag0) = 1 → bag0 = ∅ ++ bag0.max ⊦ (m = (m < bag0.max ⊃ bag0.max;m) ⊃ bag0;(bag0 -- bag0.max) ++ m) = ∅, bag0 = ∅, m < bag0.max

Rule argument

fma: m < bag0.max

Additional information

Used formulas: m < bag0.max