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Proof step 8 from proof bag-rem-subseteq (previous step)

count(n0, bag) < count(n0, bag) - (n0 = n ∧ count(n, bag) ≠ 0 ⊃ 1;0) ⊦

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 < n; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < 1 ↔ m = 0; (s)%n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n; (s)% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic