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Proof step 23 from proof subset-is-union (previous step)

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 = ∅

User applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: s0 ++ a abort: F discard: T computed: F

Additional information

Used simplifier rules: unadd: ⊦ (s ++ a) ∪ s0 = (s ∪ s0) ++ a;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;%int-empty: ⊦ s ∩ s0 = ∅ ∧ ¬ a ∈ s → s ∩ (s0 ++ a) = ∅; Used terms: s0 ++ a