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Proof step 2 from proof subsub-del (previous step)

a ∈ s0, a ∈ s, s -- a ⊆ s0 ⊦ s ⊆ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a0 ∈ s, a ∈ s0, a ∈ s, a1 ∈ s, a2 ∈ s0, ∀ a0. a0 ∈ s → a0 ∈ s0 ⊦ a0 ∈ s0

Rule argument

Additional information

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