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Proof step 3 from proof add-inter-rem (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ s1 ⊦ s1 = ∅, (s0 ++ a) ∩ s1 = s0 ∩ s1, ((s0 ++ a) ∩ s1) -- a = (s0 ∩ s1) -- a, # s1 = 0

Rule argument

Additional information

Used simplifier rules: Intersect: ⊦ a ∈ s1 ∩ s2 ↔ a ∈ s1 ∧ a ∈ s2;%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ 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%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s); from specification set-basic