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Proof step 6 from proof diffone-nodupsget (previous step)

n < # x0, a ∈ x0, ∀ n0, a1. n0 + 1 < # x0 ∧ a1 ∈ x0 → (x0 \ a1 ')[n0] = (n0 < pos(a1, x0) ⊃ x0[n0];x0[n0 + 1]) ⊦ ((a0 ' + x0) \ a ')[n] = (¬ pos(a, x0) < n ⊃ (a0 ' + x0)[n];x0[n]), x0 = [], dups(x0), a0 ∈ x0, a0 ∈ a '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n < # x0, a ∈ x0, ∀ n0, a1. n0 + 1 < # x0 ∧ a1 ∈ x0 → (x0 \ a1 ')[n0] = (n0 < pos(a1, x0) ⊃ x0[n0];x0[n0 + 1]) ⊦ (a0 ' + (x0 \ a '))[n] = (¬ pos(a, x0) < n ⊃ (a0 ' + x0)[n];x0[n]), x0 = [], dups(x0), a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: diff-n: ⊦ ¬ a ∈ y → (a ' + x) \ y = a + (x \ y);% ⊦ a ∈ b ' ↔ a = b; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)