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Proof step 9 from proof inv-isbad (previous step)

m < # pebs, mtd-cons(pebs), ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪if err0 = ESUCCESS then isbad := pebs[m].bad else isbad := ?⟫ true

Heuristic conditional right split applied rule if right

And obtained the following new goals

Proof step 10:
err0 = ESUCCESS, m < # pebs, mtd-cons(pebs), ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪isbad := pebs[m].bad⟫ true
Proof step 11:
m < # pebs, mtd-cons(pebs), ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪isbad := ?⟫ true, err0 = ESUCCESS

Rule argument

pos: right 1

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