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Proof step 6 from proof idset-idexpt (previous step)

# apebs1 = # apebs0, # apebs2 = # apebs0, ∀ m. m < # apebs0 ∧ 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

User applied rule case distinction

And obtained the following new goals

Proof step 7:
# apebs2 = # apebs0, # apebs1 = # apebs0, ∀ m. m < # apebs0 ∧ m ≠ n → apebs1[m] = apebs2[m], ∀ n0. n0 ∈ ns ∧ n0 < # apebs0 → apebs1[n0] = apebs0[n0] ⊦ n ∈ ns, ∀ n0. n0 ∈ ns ∧ n0 < # apebs0 → apebs2[n0] = apebs0[n0]
Proof step 11:
# apebs2 = # apebs0, # apebs1 = # apebs0, ∀ m. m < # apebs0 ∧ m ≠ n → apebs1[m] = apebs2[m], ∀ n0. n0 ∈ ns ∧ n0 < # apebs0 → apebs2[n0] = apebs0[n0] ⊦ n ∈ ns, ∀ n0. n0 ∈ ns ∧ n0 < # apebs0 → apebs1[n0] = apebs0[n0]

Rule argument

pos: right 1

Proof step 6 from proof idset-idexpt (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information