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Proof step 5 from proof diffone-nodupsget (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n < # x0, a ∈ x0, ∀ a1, n0. a1 ∈ x0 ∧ n0 + 1 < # x0 → (x0 \ a1 ')[n0] = (n0 < pos(a1, x0) ⊃ x0[n0];x0[n0 + 1]) ⊦ x0 = [], (a0 ' + (x0 \ a '))[n] = (¬ pos(a, x0) < n ⊃ (a0 ' + x0)[n];x0[n]), dups(x0), a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: diff-n: ⊦ ¬ a ∈ y → (a ' + x) \ y = a + (x \ y);%in-17: ⊦ a ∈ b ' ↔ a = b; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list