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Proof step 31 from proof ADD-lemma (previous step)

s0 ≠ ∅, s0 ⊆ S, S3 = V0 ∪ (S \ s0), S2 = N0 ∪ (S \ (s0 ∪ V0)), s0 = s ⊦ ⟪choose w with w ∈ s0 in { s0 := s0 -- w ; if ¬ w ∈ S3 then S3 := S3 ++ w, S2 := S2 ++ w }⟫ (s0 ⊆ S ∧ S3 = V0 ∪ (S \ s0) ∧ S2 = N0 ∪ (S \ (s0 ∪ V0)) ∧ # s0 < # s)

Heuristic symbolic execution applied rule choose right

And obtained the following new goals

Proof step 9:
s0 ≠ ∅, s0 ⊆ S, S3 = V0 ∪ (S \ s0), S2 = N0 ∪ (S \ (s0 ∪ V0)), s0 = s, ∀ w. ¬ w ∈ s0 ⊦
Proof step 30:
s0 ≠ ∅, s0 ⊆ S, S3 = V0 ∪ (S \ s0), S2 = N0 ∪ (S \ (s0 ∪ V0)), s0 = s, w ∈ s0 ⊦ ⟪s0 := s0 -- w⟫ ⟪if ¬ w ∈ S3 then S3 := S3 ++ w, S2 := S2 ++ w⟫ (s0 ⊆ S ∧ S3 = V0 ∪ (S \ s0) ∧ S2 = N0 ∪ (S \ (s0 ∪ V0)) ∧ # s0 < # s)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]