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Proof step 2 from proof Diff-ext (previous step)

∀ a, st. a ∈ st \ s0 ↔ a ∈ st ∧ ¬ a ∈ s0 ⊦ a ∈ st \ (s0 ++ a0) ↔ a ∈ st ∧ ¬ a ∈ s0 ++ a0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a, st. a ∈ st \ s0 ↔ a ∈ st ∧ ¬ a ∈ s0 ⊦ a ∈ (st \ s0) -- a0 ↔ a ∈ st ∧ a ≠ a0 ∧ ¬ a ∈ s0

Rule argument

Additional information

Used simplifier rules: Diff-rec: ⊦ st \ (s ++ a) = (st \ s) -- a;%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s); from specification set-basic