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Proof step 3 from proof sub-del-04 (previous step)

a ∈ s ⊦ (∀ a0. a0 ∈ s → a0 ≠ a ∧ a0 ∈ s0) ↔ (∀ a. a ∈ s → a ∈ s0) ∧ ¬ a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ s, a0 ∈ s, ∀ a0. a0 ∈ s → a0 ≠ a ∧ a0 ∈ s0 ⊦

Rule argument

Additional information

Used simplifier rules: ext: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);%In-empty: ⊦ ¬ a ∈ ∅; from specification infset-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification infset-basic