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Proof step 25 from proof ADD-lemma (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
w ∈ S, w ∈ V0, w ∈ s, s ⊆ S ⊦ # S = 0, V0 = ∅, # s = 0, a ∈ V0 ∪ (S \ s) ↔ a ∈ V0 ∪ (S \ (s -- w)), s = ∅, # V0 = 0, S = ∅

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]