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Proof step 1 from proof sub-del (conclusion)

⊦ s ⊆ s0 → s -- a ⊆ s0 -- a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a0 ∈ s, a1 ∈ s, ∀ a. a ∈ s → a ∈ s0 ⊦ a0 = a, a0 ∈ s0

Rule argument

Additional information

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