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Proof step 1 from proof rmpos-get (conclusion)

⊦ n + 1 < # x ∧ m < # x → (x -1l m)[n] = (n < m ⊃ x[n];x[n + 1])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
m < # x, n + 1 < # x ⊦ x = [], (x -1l m)[n] = (n < m ⊃ x[n];x.rest[n])

Rule argument

Additional information

Used simplifier rules: get-rec-01: ⊦ x ≠ [] → x[n + 1] = x.rest[n];%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1