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Proof step 19 from proof ADD-lemma (previous step)

(N0 ∪ (S \ (s ∪ V0))) ++ w ≠ N0 ∪ (S \ ((s -- w) ∪ V0)), s ⊆ S, w ∈ s, w ∈ S ⊦ S = ∅, s = ∅, # s = 0, # S = 0, w ∈ V0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
w ∈ S, w ∈ s, s ⊆ S ⊦ (N0 ∪ (S \ (s ∪ V0))) ++ w = N0 ∪ (S \ ((s -- w) ∪ V0)), # S = 0, # s = 0, s = ∅, S = ∅, w ∈ V0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]