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Proof step 25 from proof max-rem-notsame (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
m < bag0.max, bag0 = ∅ ++ bag0.max ∨ bag0.max = (bag0 -- bag0.max).max ∨ count(bag0.max, bag0) = 1 ⊦ bag0.max = (m = bag0.max ⊃ bag0;(bag0 -- bag0.max) ++ m).max, (m = bag0.max ⊃ bag0;(bag0 -- bag0.max) ++ m) = ∅, bag0 = ∅, count(bag0.max, bag0 ++ m) = 1

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic