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Proof step 2 from proof int-notempty-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a0 ∈ s0, a0 ∈ s, a1 ∈ s0, a2 ∈ s, ∀ a1. ¬ a1 ∈ s ∨ ¬ a1 ∈ s0 ⊦ a0 = a, b ∈ s

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;%int-inserte: ⊦ a ∈ s → s ∩ (s0 ++ a) ≠ ∅;%Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s)% ⊦ ¬ a ∈ ∅; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)