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Proof step 3 from proof subset-plusminus (previous step)

⊦ count(a, bag1) ≠ 0 → ((∀ a0. count(a0, bag0) ≤ count(a0, bag1 -- a)) ↔ bag0 ++ a ⊆ bag1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ count(a, bag1) = 0, (∀ a0. count(a0, bag0) ≤ count(a0, bag1) - (a0 = a ⊃ 1;0)) ↔ bag0 ++ a ⊆ bag1

Rule argument

Additional information

Used simplifier rules: count-del: ⊦ count(a, bag -- b) = count(a, bag) - (a = b ∧ count(a, bag) ≠ 0 ⊃ 1;0);