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Proof step 11 from proof ADD-lemma (previous step)

w ∈ S, w ∈ s, s ⊆ S ⊦ # S = 0, # s = 0, s = ∅, S = ∅, a = w ∨ a ∈ V0 ∪ (S \ s) ↔ a ∈ V0 ∪ (S \ (s -- w)), w ∈ V0

User applied rule case distinction

And closed the goal!

Rule argument

[CASEDARG : pos: right 5, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ a ∈ s -- b ↔ ¬ ¬ (a ≠ b ∧ a ∈ s)], [CSIMPSEQ : ⊦ a ∈ s1 ∪ s2 ↔ a ∈ s1 ∨ a ∈ s2], [CSIMPSEQ : ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1], [CSIMPSEQ : ⊦ s ∪ s0 = s0 ∪ s], [CSIMPSEQ : ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2], [CSIMPSEQ : ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s])]]