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Proof step 1 from proof int-ins (conclusion)

⊦ s ∩ (s0 ++ a) = ∅ ↔ s ∩ s0 = ∅ ∧ ¬ a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (∀ a0. ¬ a0 ∈ s ∨ a0 ≠ a ∧ ¬ a0 ∈ s0) ↔ (∀ a0. ¬ a0 ∈ s ∨ ¬ a0 ∈ s0) ∧ ¬ a ∈ s

Rule argument

Additional information

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