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Proof step 3 from proof sub-add-03 (previous step)

a = a ∨ a ∈ s → a ∈ s0, ∀ a0. a0 ∈ s → a0 ∈ s0 ⊦ (a0 ∈ s → a0 ∈ s0) ∧ a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);