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Proof step 2 from proof abs-size-apebs1 (previous step)

abs(pebs, apebs) ⊦ # pebs = # apebs ↔ true

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: abs: ⊦ abs(pebs, apebs) ↔ # pebs = # apebs ∧ (∀ n. n < # pebs → abs-peb(pebs[n], apebs[n]));