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Proof step 1 from proof int-notempty (conclusion)

⊦ s ∩ s0 = ∅ → (s ∩ (s0 ++ b) ≠ ∅ ↔ b ∈ s)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ s0, a ∈ s, a0 ∈ s0, a1 ∈ s, ∀ a0. ¬ a0 ∈ s ∨ ¬ a0 ∈ s0 ⊦ b ∈ 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;%int-ins: ⊦ s ∩ (s0 ++ a) = ∅ ↔ s ∩ s0 = ∅ ∧ ¬ a ∈ s;%int-inserte: ⊦ a ∈ s → s ∩ (s0 ++ a) ≠ ∅;%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification infset-basic%In-empty: ⊦ ¬ a ∈ ∅; from specification infset-basic