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Proof step 9 from proof crash-neutral-recover (
previous step
)
ebm-cons(avols0), ebm-cons(avols), avols ⊆ avols0 ⊦ ⟨avols := ?⟩ (avols ⊆ avols0)
Heuristic symbolic execution applied rule assign right
And obtained the following new goals
Proof step 10
:
ebm-cons(avols1), ebm-cons(avols0), avols1 ⊆ avols0 ⊦ ∃ avols. avols ⊆ avols0
Rule argument
Additional information