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Proof step 3 from proof union-subset-inj (previous step)

⊦ (∀ a. count(a, bag ∪ bag1) ≤ count(a, bag ∪ bag2)) ↔ bag1 ⊆ bag2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (∀ a. count(a, bag1) ≤ count(a, bag2)) ↔ bag1 ⊆ bag2

Rule argument

Additional information

Used simplifier rules: union: ⊦ count(a, bag0 ∪ bag1) = count(a, bag0) + count(a, bag1);%union-ass: ⊦ bag0 ∪ bag1 ∪ bag2 = (bag0 ∪ bag1) ∪ bag2;%union-com: ⊦ bag0 ∪ bag1 = bag1 ∪ bag0;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)