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Proof step 12 from proof ADD-lemma (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
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

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ a ∈ s ++ b ↔ ¬ ¬ (a = b ∨ a ∈ s)])]]