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Proof step 21 from proof difference (previous step)

count(a0, bag0) - (count(a0, bag0) ≠ 0 ⊃ 1;0) = count(a0, (bag0 -- a0) \ bag) + count(a0, bag), count(a0, bag0) ≤ count(a0, bag) + 1, count(a0, bag0) ≤ count(a0, (bag0 -- a0) \ bag) + count(a0, bag) ⊦ count(a0, (bag0 -- a0) \ bag) = 0

Heuristic cut applied rule cut formula

And closed the goal!

Rule argument

fma: count(a0, bag0) = 0

Additional information

Used simplifier rules: del-notin: ⊦ count(a, bag) = 0 → bag -- a = bag;%leprf: ⊦ m ≤ m - 1 → m = 0; (forward) from specification nat% ⊦ m - 0 = m; (s)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used formulas: count(a0, bag0) = 0