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Proof step 14 from proof diff-add-sw (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
count(a0, bag0) = n0 + count(a0, bag1), count(a, bag1) = 0 ⊦ a0 = a, n0 = count(a0, bag0) - count(a0, bag1)

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-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ m + n < m; (s)% ⊦ m < n → n ≠ 0; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ a ≠ b → count(a, bag ++ b) = count(a, bag); (s)