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Proof step 11 from proof subset-plusminus (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
count(a, bag1) ≤ count(a, bag0), ∀ a0. count(a0, bag0) ≤ count(a0, bag1) - (a0 = a ⊃ 1;0) ⊦ count(a, bag1) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ count(a, bag ++ a) = count(a, bag) + 1; (s)