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Proof step 14 from proof ADD-lemma (previous step)

(V0 ∪ (S \ s)) ++ w ≠ V0 ∪ (S \ (s -- w)), 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 13:
w ∈ S, w ∈ s, s ⊆ S ⊦ (V0 ∪ (S \ s)) ++ w = V0 ∪ (S \ (s -- w)), # S = 0, # s = 0, s = ∅, S = ∅, w ∈ V0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]