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Proof step 1 from proof undel-04 (conclusion)

⊦ (s -- a) ∪ (s0 -- a) = (s ∪ s0) -- a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a0 ≠ a ∧ a0 ∈ s ∨ a0 ≠ a ∧ a0 ∈ s0 ⊦ a0 ≠ a ∧ (a0 ∈ s ∨ a0 ∈ s0)

Rule argument

Additional information

Used simplifier rules: ext: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);%unrax: ⊦ a ∈ s ∪ s0 ↔ ¬ ¬(a ∈ s ∨ a ∈ s0);%del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s);%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;