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Proof step 4 from proof idset-trans-intersect (previous step)

# apebs2 = # apebs0, # apebs1 = # apebs0, n < # apebs0, n ∈ ns0, n ∈ ns1, n ∈ ns1 ∧ n < # apebs0 → apebs1[n] = apebs0[n], ∀ n0. n0 ∈ ns0 ∧ n0 < # apebs0 → apebs2[n0] = apebs1[n0], ∀ n0. n0 ∈ ns1 ∧ n0 < # apebs0 → apebs1[n0] = apebs0[n0] ⊦ # ns0 = 0, # ns1 = 0, apebs2[n] = apebs0[n], ns0 = ∅, ns1 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# apebs1 = # apebs0, # apebs2 = # apebs0, apebs1[n] = apebs0[n], n < # apebs0, n ∈ ns1, n ∈ ns0, ∀ n0. n0 ∈ ns1 ∧ n0 < # apebs0 → apebs1[n0] = apebs0[n0], ∀ n0. n0 ∈ ns0 ∧ n0 < # apebs0 → apebs2[n0] = apebs1[n0] ⊦ # ns1 = 0, # ns0 = 0, apebs2[n] = apebs0[n], ns0 = ∅, ns1 = ∅

Proof step 4 from proof idset-trans-intersect (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information