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Proof step 6 from proof inv-format (previous step)

err = err0, ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪if err = ESUCCESS then avols := ∅ else avols := ?⟫ (err = ESUCCESS → ebm-cons(avols))

Heuristic conditional right split applied rule if right

And obtained the following new goals

Proof step 7:
err = ESUCCESS, err = err0, ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪avols := ∅⟫ (err = ESUCCESS → ebm-cons(avols))
Proof step 11:
err = err0, ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪avols := ?⟫ (err = ESUCCESS → ebm-cons(avols)), err = ESUCCESS

Rule argument

pos: right 1

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