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Proof step 35 from proof ADD-lemma (conclusion)

N = N0, V = V0 ⊦ ⟪ADD(S; V, N)⟫ (N = N0 ∪ (S \ V0) ∧ V = V0 ∪ S)

Heuristic weak unfold applied rule call right

And obtained the following new goals

Proof step 1:
⊦
Proof step 34:
N = N0, V = V0 ⊦ ⟪S1 := N, S0 := V, s := S⟫ ⟪while s ≠ ∅ do choose w with w ∈ s in { s := s -- w ; if ¬ w ∈ S0 then S0 := S0 ++ w, S1 := S1 ++ w }⟫ (S1 = N0 ∪ (S \ V0) ∧ S0 = V0 ∪ S)

Rule argument

[fmapos-arg: pos: right 1]

Additional information

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