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Proof step 3 from proof free-known (previous step)

(r => o * P)(H) ⊦ ⟪H0 := H⟫ ⟪if ¬ r ∈ H0 then abort else H0 := H0 -- r⟫ P(H0)

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 2:
(r => o * P)(H), H1 = H ⊦ ⟪if ¬ r ∈ H1 then abort else H1 := H1 -- r⟫ P(H1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]