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Proof step 1 from proof intersection-shift (conclusion)

⊦ a ∈ s → s ∩ (s0 ++ a) = (s ∩ s0) ++ a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ s, a1 ∈ s, a0 = a ∨ a0 ∈ s ∧ a0 ∈ s0 ⊦ a0 ∈ s ∧ (a0 = a ∨ a0 ∈ s0)

Rule argument

Additional information

Used simplifier rules: int-rax: ⊦ a ∈ s1 ∩ s2 ↔ ¬ ¬(a ∈ s1 ∧ a ∈ s2);%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;%Intersect: ⊦ a ∈ s1 ∩ s2 ↔ a ∈ s1 ∧ a ∈ s2;%In-empty: ⊦ ¬ a ∈ ∅; 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%Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); from specification set-basic