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Proof step 1 from proof insdel-sub (conclusion)

⊦ ¬ s -- a ⊆ s0 ∧ a ∈ s → ¬ s ⊆ s0 ++ a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;%In-empty: ⊦ ¬ a ∈ ∅; from specification set-basic%Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s); from specification set-basic%Subset: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2); from specification set-basic