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Proof step 21 from proof subset-is-union (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 22:
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 = ∅

Rule argument

pos: left 2 terms: s abort: F discard: F computed: T

Additional information

Used terms: s