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Proof step 7 from proof intu-empty (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a0 ∈ s1, a0 ∈ s, a ∈ s1, a1 ∈ s, ∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0 ∧ ¬ a ∈ s1 ⊦

Rule argument

Additional information

Used simplifier rules: unrax: ⊦ a ∈ s ∪ s0 ↔ ¬ ¬(a ∈ s ∨ a ∈ s0);%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%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