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Proof step 12 from proof rmpos-get (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
m < # x0, m < m1, m1 < # x0, m + 1 < # x0, ∀ m0, n0. m0 < # x0 ∧ n0 + 1 < # x0 → (x0 -1l m0)[n0] = (n0 < m0 ⊃ x0[n0];x0.rest[n0]) ⊦ m1 = 0, (x0 -1l m1)[m] = x0[m], x0 = []

Rule argument

Additional information

Used simplifier rules: get-succ: ⊦ (a ' + x)[n + 1] = x[n];%succ: ⊦ m +1 = m + 1; from specification nat-basic2%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2