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Proof step 1 from proof subsub-del (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a0 ∈ s, a ∈ s, a ∈ s0, a1 ∈ s, a2 ∈ s0, ∀ a0. a0 ∈ s → a0 ∈ s0 ⊦ a0 ∈ s0

Rule argument

Additional information

Used simplifier rules: del-rax: ⊦ 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%Subset: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2); from specification set-basic