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Proof step 11 from proof intu-empty (previous step)

a ∈ s0, a ∈ s, a0 ∈ s, ∀ a. ¬ a ∈ s ∨ ¬ a ∈ s1, ∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
a0 ∈ s, a ∈ s, a ∈ s0, a1 ∈ s0, ∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0, ∀ a. ¬ a ∈ s ∨ ¬ a ∈ s1 ⊦

Rule argument

Additional information

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