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Proof step 24 from proof subset-is-union (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
s ⊆ s1, a ∈ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ s1 = ∅, # s1 = 0, ∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic