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Proof step 5 from proof int-ins (previous step)

∀ a0. ¬ a0 ∈ s ∨ a0 ≠ a ∧ ¬ a0 ∈ s0 ⊦ ¬ a0 ∈ s ∨ ¬ a0 ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a0 ∈ s0, a0 ∈ s, a1 ∈ s0, a2 ∈ s, ∀ a0. ¬ a0 ∈ s ∨ a0 ≠ a ∧ ¬ a0 ∈ s0 ⊦

Rule argument

Additional information

Used simplifier rules: ext: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);%In-empty: ⊦ ¬ a ∈ ∅; from specification infset-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification infset-basic