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Proof step 2 from proof idset-idexpt (previous step)

id○(apebs1, apebs2, n) ⊦ id(apebs1, apebs0, ns) ↔ id(apebs2, apebs0, ns), n ∈ ns

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# apebs2 = # apebs0, # apebs1 = # apebs0, id○(apebs1, apebs2, n) ⊦ (∀ n0. n0 ∈ ns ∧ n0 < # apebs0 → apebs1[n0] = apebs0[n0]) ↔ (∀ n0. n0 ∈ ns ∧ n0 < # apebs0 → apebs2[n0] = apebs0[n0]), n ∈ ns

Rule argument

Additional information

Used simplifier rules: id-set: ⊦ id(apebs0, apebs1, ns) ↔ # apebs0 = # apebs1 ∧ (∀ n. n ∈ ns ∧ n < # apebs0 ∧ n < # apebs1 → apebs0[n] = apebs1[n]);%idexcept-size: ⊦ id○(apebs, apebs0, n) → # apebs = # apebs0; (forward) from specification array-copy[apebs]