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Proof step 10 from proof subset-is-union (previous step)

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 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
(s -- a) ∩ s0 = ∅, a ∈ s ⊦ s = ∅, # s = 0, a ∈ s0, ∃ s1. s ∪ s1 = s0 ∪ s ∧ s ∩ s1 = ∅

Rule argument

Additional information

Used simplifier rules: insdel-sub: ⊦ ¬ s -- a ⊆ s0 ∧ a ∈ s → ¬ s ⊆ s0 ++ a;%unsub: ⊦ s ⊆ s ∪ s0;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%unin-01: ⊦ ¬ a ∈ s ∪ s0 ↔ ¬ a ∈ s ∧ ¬ a ∈ s0;%Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;%undel-03: ⊦ ¬ a ∈ s → s ∪ (s0 -- a) = (s ∪ s0) -- a;%delete-insert: ⊦ (s -- a) ++ a = s ++ a;%uninsert: ⊦ a ∈ s → (s ∪ s0) ++ a = s ∪ s0;