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Proof step 3 from proof fail-prepost (previous step)

⊦ ⟪choose err0 with ⌊err0⌋ in err := err0⟫ ⌊err⌋

Heuristic symbolic execution applied rule choose right

And obtained the following new goals

Proof step 4:
⌊err0⌋ ⊦ ⟪err := err0⟫ ⌊err⌋
Proof step 5:
⊦ ∃ err0. ⌊err0⌋

Rule argument

Additional information