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Proof step 28 from proof ADD-lemma (previous step)

s0 ≠ ∅, s0 ⊆ S, S3 = V0 ∪ (S \ s0), S2 = N0 ∪ (S \ (s0 ∪ V0)), s0 = s, w ∈ s0, ¬ ¬ w ∈ S3 ⊦ ⟪skip⟫ (s0 -- w ⊆ S ∧ S3 = V0 ∪ (S \ (s0 -- w)) ∧ S2 = N0 ∪ (S \ ((s0 -- w) ∪ V0)) ∧ # (s0 -- w) < # s)

Heuristic symbolic execution applied rule skip right

And obtained the following new goals

Proof step 27:
s0 ≠ ∅, s0 ⊆ S, S3 = V0 ∪ (S \ s0), S2 = N0 ∪ (S \ (s0 ∪ V0)), s0 = s, w ∈ s0, ¬ ¬ w ∈ S3, ¬ s0 -- w ⊆ S ∨ S3 ≠ V0 ∪ (S \ (s0 -- w)) ∨ S2 ≠ N0 ∪ (S \ ((s0 -- w) ∪ V0)) ∨ ¬ # (s0 -- w) < # s ⊦

Rule argument

[emptyarg]

Additional information

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