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Proof step 16 from proof subset-plusminus (previous step)

a0 = a, count(a0, bag1) - 1 < count(a0, bag0), ∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1) ⊦ count(a, bag1) = 0, count(a0, bag0) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
count(a, bag1) ≤ count(a, bag0), ∀ a0. count(a0, bag0 ++ a) ≤ count(a0, bag1) ⊦ count(a, bag1) = 0, count(a, bag0) = 0

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ 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%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%ls-01: n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat