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Proof step 10 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, count(a0, bag0 ++ a) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
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: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%count-same: ⊦ count(a, bag ++ a) = count(a, bag) + 1; from specification bag-basic