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Proof step 5 from proof union-diff-distr (previous step)

count(a, bag0) = n0 + count(a, bag1) ∧ count(a, bag1) ≤ count(a, bag0), ∀ a. count(a, bag1) ≤ count(a, bag0) ⊦ n0 + count(a, bag2) = (count(a, bag1) ≤ count(a, bag0) + count(a, bag2) ⊃ count(a, bag0) + count(a, bag2) - count(a, bag1);0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat