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Proof step 24 from proof rmpos-in (previous step)

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

User applied rule constructor cut

And obtained the following new goals

Proof step 25:
n = n1 +1, a ∈ (a0 ' + x0) -1l n, n ≤ # x0, ∀ n0, a1. n0 < # x0 → (a1 ∈ x0 -1l n0 ↔ a1 ≠ x0[n0] ∧ a1 ∈ x0) ⊦ a = a0, x0 = [], a ∈ x0, a0 ∈ x0, dups(x0)

Rule argument

var: n

Additional information

Used simplifier rules: delpos-base: ⊦ (a ' + x) -1l 0 = x;%nat freely generated by 0, +1;;%a: ⊦ (x + y) + z = x + y + z; from specification list