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Proof step 2 from proof uninsubdel (previous step)

a ∈ (s1 -- b) ∪ s0, s1 ⊆ s ⊦ a ∈ s ∪ s0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Union: ⊦ a ∈ s1 ∪ s2 ↔ a ∈ s1 ∨ a ∈ s2;%Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%sub-in: ⊦ a ∈ s ∧ s ⊆ s0 → a ∈ s0; (forward) from specification set-basic