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Proof step 22 from proof subset-is-union (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
s ∩ s0 = ∅, ∀ s2. s2 ⊆ s ∪ s0 → (∃ s3. s2 ∪ s3 = s ∪ s0 ∧ s2 ∩ s3 = ∅) ⊦ a ∈ s, a ∈ s0, ∃ s2. s ∪ s2 = (s ∪ s0) ++ a ∧ s ∩ s2 = ∅

Rule argument

Additional information

Used simplifier rules: unsub: ⊦ s ⊆ s ∪ s0;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%unin: ⊦ ¬ a ∈ s → (a ∈ s ∪ s0 ↔ a ∈ s0);