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Proof step 2 from proof difference (previous step)

count(a, bag0 \ bag1) = (count(a, bag1) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag1);0) ⊦ count(a, bag0 \ bag1) = (count(a, bag1) ≤ count(a, bag0) ⊃ count(a, bag0) - count(a, bag1);0)

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%zero: ⊦ m - m = 0; from specification nat%lels-03: ⊦ m < n → m ≤ n; from specification nat