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Proof step 5 from proof rmpos-in (previous step)

n ≤ # x0, ∀ n0, a1. n0 < # x0 → (a1 ∈ x0 -1l n0 ↔ a1 ≠ x0[n0] ∧ a1 ∈ x0) ∨ x0 = [] ⊦ x0 = [], a ∈ (a0 ' + x0) -1l n ↔ a ≠ (a0 ' + x0)[n] ∧ (a = a0 ∨ a ∈ x0), a0 ∈ x0, dups(x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n ≤ # x0, ∀ n0, a1. n0 < # x0 → (a1 ∈ x0 -1l n0 ↔ a1 ≠ x0[n0] ∧ a1 ∈ x0) ⊦ x0 = [], a ∈ (a0 ' + x0) -1l n ↔ a ≠ (a0 ' + x0)[n] ∧ (a = a0 ∨ a ∈ x0), a0 ∈ x0, dups(x0)

Proof step 5 from proof rmpos-in (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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