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Proof step 11 from proof inv-erase (previous step)

pebs[m] = mkpeb(buf, m0, false), m < # pebs, mtd-cons(pebs), ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪if err0 = ESUCCESS then pebs[m] := mkpeb(emptypeb, 0, pebs[m].bad)⟫ mtd-cons(pebs)

Heuristic conditional right split applied rule if right

And obtained the following new goals

Proof step 12:
pebs[m] = mkpeb(buf, m0, false), err0 = ESUCCESS, m < # pebs, mtd-cons(pebs), ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪pebs[m] := mkpeb(emptypeb, 0, pebs[m].bad)⟫ mtd-cons(pebs)
Proof step 18:
pebs[m] = mkpeb(buf, m0, false), m < # pebs, mtd-cons(pebs), ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪skip⟫ mtd-cons(pebs), err0 = ESUCCESS

Rule argument

pos: right 1

Proof step 11 from proof inv-erase (previous step)

Heuristic conditional right split applied rule if right

And obtained the following new goals

Rule argument

Additional information