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Proof step 1 from proof mtd-cons-extract (conclusion)
⊦ n < # pebs → (mtd-cons(pebs) ↔ mtd-valid-peb(pebs[n]) ∧ mtd-cons(pebs))
System applied rule normalize
And obtained the following new goals
Proof step 2
:
n < # pebs ⊦ mtd-cons(pebs) ↔ mtd-valid-peb(pebs[n]) ∧ mtd-cons(pebs)
Rule argument
Additional information