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Proof step 24 from proof rmpos-get (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
m + 1 < # x0, m1 + 1 < # x0, m1 < # x0, m1 ≤ m, ∀ n0, m0. n0 + 1 < # x0 ∧ m0 < # x0 → (x0 -p m0)[n0] = (n0 < m0 ⊃ x0[n0];x0.tail[n0]) ⊦ (x0 -p m1)[m] = x0.tail[m], x0 = []

Rule argument

Additional information

Used simplifier rules: get-rec-01: ⊦ x ≠ [] → x[n + 1] = x.tail[n];% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)