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Proof step 21 from proof inv-format (previous step)

avols = ∅, err = err0, err = ESUCCESS, ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ default-volid ∈ avols[default-volid, fill(mkabstractvolume(n), erased)] ∧ ebm-cons(avols[default-volid, fill(mkabstractvolume(n), erased)])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
⊦ ebm-cons((∅)[default-volid, fill(mkabstractvolume(n), erased)])

Rule argument

Additional information

Used simplifier rules: ⊦ v ∈ avols[v, evol]; (s)% ⊦ ¬ ⌊ESUCCESS⌋ ; (s)