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Proof step 21 from proof rmpos-get (previous step)

n = m + 1, m1 < # x0, m1 < n, n < # x0, m1 + 1 < # x0, ∀ m0, n0. m0 < # x0 ∧ n0 + 1 < # x0 → (x0 -1l m0)[n0] = (n0 < m0 ⊃ x0[n0];x0.rest[n0]) ⊦ x0 = [], n = 0, (x0 -1l m1)[m] = x0[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
m + 1 < # x0, m1 + 1 < # x0, m1 < # x0, m1 ≤ m, ∀ m0, n0. m0 < # x0 ∧ n0 + 1 < # x0 → (x0 -1l m0)[n0] = (n0 < m0 ⊃ x0[n0];x0.rest[n0]) ⊦ x0 = [], (x0 -1l m1)[m] = x0.rest[m]

Rule argument

Additional information

Used simplifier rules: get-rec-01: ⊦ x ≠ [] → x[n + 1] = x.rest[n];%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%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