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Proof step 7 from proof inv-contains (previous step)

boolvar ↔ boolvar1, boolvar0 ↔ boolvar2, ro.inodes?, fo.inodes?, log-cons(log, fs), ri-inj(ri), ri-inj(fi), ⌊err0⌋ ∨ err0 = ESUCCESS ⊦ ⟪if err0 = ESUCCESS then boolvar := key ∈ ri else boolvar := ?⟫ true

Heuristic conditional right split applied rule if right

And obtained the following new goals

Rule argument

pos: right 1

Additional information