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Proof step 2 from proof list2set-append (previous step)

∀ y. list2set(x0 + y) = list2set(x0) ∪ list2set(y) ⊦ list2set((a + x0) + y) = list2set(a + x0) ∪ list2set(y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ y. list2set(x0 + y) = list2set(x0) ∪ list2set(y) ⊦ list2set(x0 + y) ++ a = (list2set(x0) ∪ list2set(y)) ++ a

Rule argument

Additional information

Used simplifier rules: list2set-rec: ⊦ list2set(a ' + x) = list2set(x) ++ a;% ⊦ s ∪ s0 = s0 ∪ s;% ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;% ⊦ (s ++ a) ∪ s0 = (s ∪ s0) ++ a;% ⊦ (x + y) + z = x + y + z;% ⊦ a + x = a ' + x;