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Proof step 24 from proof bag-rem-empty (previous step)

count(n1, bag) - (n1 = n ⊃ 1;0) = 0, count(n, bag) = 1, ∀ n2. count(n2, bag -- n) = 0 ⊦ count(n1, bag) = count(n1, ∅ ++ n)

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ m - 0 = m; (s)% ⊦ n ≠ n0 → count(n, bag0 ++ n0) = count(n, bag0); (s)% ⊦ count(n, ∅) = 0; (s)% ⊦ count(n, ∅ ++ n) = 1; (s)