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

# apebs2 = # apebs0, # apebs1 = # apebs0, n0 < # apebs0, n0 ∈ ns, ∀ m. m < # apebs0 ∧ m ≠ n → apebs1[m] = apebs0[m], ∀ n1. n1 ∈ ns ∧ n1 < # apebs0 → apebs2[n1] = apebs1[n1] ⊦ # ns = 0, apebs2[n0] = apebs0[n0], n0 = n, ns = ∅

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 7:
# apebs1 = # apebs0, # apebs2 = # apebs0, n0 < # apebs0, n0 ∈ ns, n0 ∈ ns ∧ n0 < # apebs0 → apebs2[n0] = apebs1[n0], ∀ n1. n1 ∈ ns ∧ n1 < # apebs0 → apebs2[n1] = apebs1[n1], ∀ m. m < # apebs0 ∧ m ≠ n → apebs1[m] = apebs0[m] ⊦ # ns = 0, apebs2[n0] = apebs0[n0], n0 = n, ns = ∅

Rule argument

pos: left 6 terms: n0

Additional information

Used terms: n0