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Proof step 1 from proof diff-eq-diff-eq-union (conclusion)
⊦ as2 \ as1 = as3 ∧ as1 \ as0 = as4 ∧ as1 ⊆ as2 ∧ as0 ⊆ as1 → as2 \ as0 = as3 ∪ as4
System applied rule normalize
And obtained the following new goals
Proof step 2
:
as2 \ as1 = as3, as1 \ as0 = as4, as1 ⊆ as2, as0 ⊆ as1 ⊦ as2 \ as0 = as3 ∪ as4
Rule argument
Additional information