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

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

Heuristic cut applied rule cut formula

And obtained the following new goals

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

Rule argument

fma: x0 = []

Additional information

Used simplifier rules: delnr-one: ⊦ a ' -p 0 = [];%one-03: ⊦ a '[0] = a;% ⊦ x + [] = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ ¬ a ∈ []; (s)% ⊦ # [] = 0; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ ¬ dups([]); (s)% ⊦ ¬ m < 0; (s) Used formulas: x0 = []