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Proof step 3 from proof subset-is-union (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ a ∈ s → s ++ a = s
Proof step 5:
¬ a ∈ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅), s ⊆ s1 ++ a ⊦ ∃ s2. s ∪ s2 = s1 ++ a ∧ s ∩ s2 = ∅
Proof step 24:
a ∈ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅), s ⊆ s1 ⊦ ∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅

Rule argument

spec: set-basic inst: name: eq seq: ⊦ a ∈ s → s ++ a = s subst: [a, s] → [a, s1] flags: rewrite

Additional information

Used terms: a, s1