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Proof step 5 from proof intu-empty (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ∈ s, a ∈ s0, a0 ∈ s, a1 ∈ s0, ∀ a0. ¬ a0 ∈ s ∨ ¬ a0 ∈ s0 ∧ ¬ a0 ∈ 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;% ⊦ ¬ a ∈ ∅; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification infset-basic