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Proof step 13 from proof rmpos-get (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
m < # x0, m < m1, m1 < # x0, m + 1 < # x0, ∀ n0, m0. n0 + 1 < # x0 ∧ m0 < # x0 → (x0 -p m0)[n0] = (n0 < m0 ⊃ x0[n0];x0.tail[n0]) ⊦ (x0 -p m1)[m] = x0[m], x0 = []

Rule argument

Additional information

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