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Proof step 6 from proof subset-is-union (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ ¬ a ∈ s → (s ⊆ s0 ++ a ↔ s ⊆ s0)
Proof step 8:
¬ ¬ a ∈ s, s ⊆ s1 ++ a, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ a ∈ s1, ∃ s2. s ∪ s2 = s1 ++ a ∧ s ∩ s2 = ∅
Proof step 20:
¬ a ∈ s, s ⊆ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ a ∈ s1, ∃ s2. s ∪ s2 = s1 ++ a ∧ s ∩ s2 = ∅

Rule argument

spec: set-basic inst: name: sub-insert seq: ⊦ ¬ a ∈ s → (s ⊆ s0 ++ a ↔ s ⊆ s0) subst: [a, s0, s] → [a, s1, s] flags: rewrite

Additional information

Used terms: a, s1, s