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Proof step 2 from proof rem-union-ins (previous step)

a ∈ s1 ⊦ (s0 ∪ (s1 -- a)) ++ a = (s0 ∪ s1) ++ a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a ∈ s1 ⊦ s1 = ∅, (s0 ∪ (s1 -- a)) ++ a = s0 ∪ s1, # s1 = 0

Rule argument

Additional information

Used simplifier rules: uninsert: ⊦ a ∈ s → (s ∪ s0) ++ a = s ∪ s0;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic