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Proof step 20 from proof ADD-lemma (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 14:
(V0 ∪ (S \ s)) ++ w ≠ V0 ∪ (S \ (s -- w)), s ⊆ S, w ∈ s, w ∈ S ⊦ S = ∅, s = ∅, # s = 0, # S = 0, w ∈ V0
Proof step 19:
(N0 ∪ (S \ (s ∪ V0))) ++ w ≠ N0 ∪ (S \ ((s -- w) ∪ V0)), s ⊆ S, w ∈ s, w ∈ S ⊦ S = ∅, s = ∅, # s = 0, # S = 0, w ∈ V0

Rule argument

[CASEDARG : pos: left 4, ()]

Additional information

[[CONCRETESIMPRULES : ()]]