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Proof step 8 from proof subset-is-union (previous step)

¬ ¬ a ∈ s, s ⊆ s1 ++ a, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ a ∈ s1, ∃ s2. s ∪ s2 = s1 ++ a ∧ s ∩ s2 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
s ⊆ s1 ++ a, a ∈ s, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ s = ∅, # s = 0, a ∈ s1, ∃ s2. s ∪ s2 = s1 ++ a ∧ s ∩ s2 = ∅

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%in-03: ⊦ ¬ a ∈ s → (a ∈ s ++ b ↔ a = b); from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%sub-in: ⊦ a ∈ s ∧ s ⊆ s0 → a ∈ s0; (forward) from specification set-basic