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Proof step 2 from proof diff-eq-diff-eq-union (previous step)

as2 \ as1 = as3, as1 \ as0 = as4, as1 ⊆ as2, as0 ⊆ as1 ⊦ as2 \ as0 = as3 ∪ as4

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
as0 ⊆ as1, as1 ⊆ as2, as0 ⊆ as2 ⊦ as2 \ as0 = (as2 \ as1) ∪ (as1 \ as0)

Rule argument

Additional information

Used simplifier rules: sub-trans: ⊦ as ⊆ as0 ∧ as0 ⊆ as1 → as ⊆ as1; (forward) from specification set-basic[node-store]