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Proof step 2 from proof intersection-shift (previous step)

a ∈ s ⊦ s ∩ (s0 ++ a) = (s ∩ s0) ++ a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a ∈ s, a1 ∈ s, a0 = a ∨ a0 ∈ s ∧ a0 ∈ s0 ⊦ a0 ∈ s ∧ (a0 = a ∨ a0 ∈ s0)

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;% ⊦ ¬ a ∈ ∅; (s)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)