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Proof step 1 from proof undel-01 (conclusion)

⊦ (s ∪ s0) -- a = s ∪ (s0 -- a) ↔ ¬ a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: undel: ⊦ a ∈ s → s ∪ (s0 -- a) = s ∪ s0;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%del-in-03: ⊦ s -- a = s ↔ ¬ a ∈ s;%unrax: ⊦ a ∈ s ∪ s0 ↔ ¬ ¬(a ∈ s ∨ a ∈ s0);%ext: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);%del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s);%Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;