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Proof step 1 from proof sub-del-04 (conclusion)

⊦ s ⊆ s0 -- a ↔ s ⊆ s0 ∧ ¬ a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (∀ a0. a0 ∈ s → a0 ≠ a ∧ a0 ∈ s0) ↔ (∀ a. a ∈ s → a ∈ s0) ∧ ¬ a ∈ s

Rule argument

Additional information

Used simplifier rules: subdef: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2);%del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s);