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Proof step 5 from proof add-inter-rem (previous step)

a ∈ s1 ⊦ a ∈ s0 ∩ s1, s1 = ∅, (s0 ++ a) ∩ s1 = s0 ∩ s1, ((s0 ++ a) ∩ s1) -- a = (s0 ∩ s1) -- a, # s1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ∈ s1 ⊦ # s1 = 0, s1 = ∅, (s0 ++ a) ∩ s1 = s0 ∩ s1, ((s0 ++ a) ∩ s1) -- a = s0 ∩ s1, a ∈ s0

Rule argument

Additional information

Used simplifier rules: del-in: ⊦ ¬ a ∈ s → s -- a = s;%Intersect: ⊦ a ∈ s1 ∩ s2 ↔ a ∈ s1 ∧ a ∈ s2;%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;