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Proof step 13 from proof diff-add-sw (previous step)

a0 ≠ a, count(a, bag1) = 0, count(a0, bag0) = n0 + count(a0, bag1) ⊦ n0 = count(a0, (bag0 \ bag1) ++ a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: difference: ⊦ count(a, bag0 \ bag1) = (count(a, bag1) ≤ count(a, bag0) ⊃ count(a, bag0) - count(a, bag1) ;0 );%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%count-other: ⊦ a ≠ b → count(a, bag ++ b) = count(a, bag); from specification bag-basic