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Proof step 2 from proof subset-is-union (previous step)

s ⊆ s0 ⊦ ∃ s1. s ∪ s1 = s0 ∧ s ∩ s1 = ∅

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅), s ⊆ s1 ++ a ⊦ ∃ s2. s ∪ s2 = s1 ++ a ∧ s ∩ s2 = ∅

Rule argument

indvar: s0

Additional information

Used simplifier rules: une-01: ⊦ s1 ∪ s2 = ∅ ↔ s1 = ∅ ∧ s2 = ∅;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%int-e: ⊦ s ∩ ∅ = ∅;%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;%set generated by ∅, ++;;%subset-empty: ⊦ s ⊆ ∅ ↔ s = ∅; from specification set-basic