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Proof step 1 from proof unsubdel (conclusion)

⊦ s ⊆ s0 ∧ s1 ⊆ s2 ∧ ¬ a ∈ s ∧ ¬ a ∈ s1 → s ∪ s1 ⊆ (s0 -- a) ∪ (s2 -- a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
∀ a. a ∈ s → a ∈ s0, ∀ a. a ∈ s1 → a ∈ s2, a0 ∈ s ∨ a0 ∈ s1 ⊦ a ∈ s1, a ∈ s, a0 ≠ a ∧ (a0 ∈ s2 ∨ a0 ∈ s0)

Rule argument

Additional information

Used simplifier rules: undel-04: ⊦ (s -- a) ∪ (s0 -- a) = (s ∪ s0) -- a;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%unrax: ⊦ a ∈ s ∪ s0 ↔ ¬ ¬(a ∈ s ∨ a ∈ s0);%del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s);%Subset: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2); from specification set-basic