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Proof step 9 from proof subset-is-union (previous step)

s ⊆ s1 ++ a, a ∈ s, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ s = ∅, # s = 0, a ∈ s1, ∃ s2. s ∪ s2 = s1 ++ a ∧ s ∩ s2 = ∅

User applied rule all left

And obtained the following new goals

Proof step 10:
s ⊆ s1 ++ a, a ∈ s, s -- a ⊆ s1 → (∃ s2. (s -- a) ∪ s2 = s1 ∧ (s -- a) ∩ s2 = ∅) ⊦ s = ∅, # s = 0, a ∈ s1, ∃ s2. s ∪ s2 = s1 ++ a ∧ s ∩ s2 = ∅

Rule argument

pos: left 3 terms: s -- a abort: F discard: T computed: F

Additional information

Used terms: s -- a