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Proof step 1 from proof in-diff-notfirst (conclusion)

⊦ ¬ a ∈ bag0 → ¬ a ∈ bag0 \ bag1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: in: ⊦ a ∈ bag ↔ count(a, bag) ≠ 0;%difference: ⊦ count(a, bag0 \ bag1) = (count(a, bag1) ≤ count(a, bag0) ⊃ count(a, bag0) - count(a, bag1) ;0 );%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat%zero-02: ⊦ m ≤ n → (n - m = 0 ↔ n = m); from specification nat