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Proof step 18 from proof subset-is-union (previous step)

∀ a0. a0 ∈ s0 ∩ s ↔ a0 ∈ ∅ ++ a, a ∈ s ⊦ # s = 0, s = ∅, a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
a ∈ s, ∀ a0. a0 ∈ s0 ∧ a0 ∈ s ↔ a0 = a ⊦ # s = 0, s = ∅, a ∈ 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;%in-04: ⊦ a ∈ ∅ ++ b ↔ a = b; from specification set-basic