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Proof step 3 from proof idset-idexpt (previous step)

# 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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ id○(apebs, apebs0, n) ↔ # apebs = # apebs0 ∧ (∀ m. m < # apebs ∧ m ≠ n → apebs[m] = apebs0[m])
Proof step 5:
# apebs1 = # apebs2, # apebs1 = # apebs0, # apebs2 = # apebs0, ∀ m. m < # apebs1 ∧ m ≠ n → apebs1[m] = apebs2[m] ⊦ (∀ n0. n0 ∈ ns ∧ n0 < # apebs0 → apebs1[n0] = apebs0[n0]) ↔ (∀ n0. n0 ∈ ns ∧ n0 < # apebs0 → apebs2[n0] = apebs0[n0]), n ∈ ns

Rule argument

spec: array-copy, inst: apebs name: Idexcept seq: ⊦ id○(apebs, apebs0, n) ↔ # apebs = # apebs0 ∧ (∀ m. m < # apebs ∧ m ≠ n → apebs[m] = apebs0[m]) subst: [n, apebs0, apebs] → [n, apebs2, apebs1] flags: <all paths>

Additional information

Used terms: n, apebs2, apebs1