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Proof step 2 from proof union-diff-distr (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 3:
∀ a. count(a, bag1) ≤ count(a, bag0) ⊦ 0 + count(a, bag2) = (count(a, bag1) ≤ count(a, bag0) + count(a, bag2) ⊃ count(a, bag0) + count(a, bag2) - count(a, bag1);0), count(a, bag1) ≤ count(a, bag0)

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ m + n < m; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)