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Proof step 19 from proof rmpos-get (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
m1 < # x0, m1 < n, n < # x0, m1 + 1 < # x0, ∀ n0, m0. n0 + 1 < # x0 ∧ m0 < # x0 → (x0 -p m0)[n0] = (n0 < m0 ⊃ x0[n0];x0.tail[n0]) ⊦ (a ' + (x0 -p m1))[n] = x0[n], x0 = []

Rule argument

Additional information

Used simplifier rules: delpos-rec: ⊦ (a ' + x) -p (n + 1) = a + (x -p n);% ⊦ 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% ⊦ a + x = a ' + x; (s)% ⊦ m +1 = m + 1; (s)