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Proof step 19 from proof diffone-nodupsget (previous step)

n = m + 1, n < # x0, a ∈ x0, n ≤ pos(a, x0), ∀ a1, n0. a1 ∈ x0 ∧ n0 + 1 < # x0 → (x0 \ a1 ')[n0] = (n0 < pos(a1, x0) ⊃ x0[n0];x0[n0 + 1]) ⊦ x0 = [], (x0 \ a ')[m] = x0[m], n = 0, dups(x0), a0 ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
m + 1 < # x0, a ∈ x0, m < pos(a, x0), ∀ a1, n0. a1 ∈ x0 ∧ n0 + 1 < # x0 → (x0 \ a1 ')[n0] = (n0 < pos(a1, x0) ⊃ x0[n0];x0[n0 + 1]) ⊦ x0 = [], (x0 \ a ')[m] = x0[m], pos(a, x0) = 0, dups(x0), a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2