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Proof step 4 from proof rmpos-get (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n < # x0, m ≤ # x0, ∀ n0, m0. n0 + 1 < # x0 ∧ m0 < # x0 → (x0 -p m0)[n0] = (n0 < m0 ⊃ x0[n0];x0.tail[n0]) ⊦ ((a ' + x0) -p m)[n] = (n < m ⊃ (a ' + x0)[n];x0[n]), x0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ 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 + n < m + n0 ↔ n < n0; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ (a ' + x).tail = x; (s)% ⊦ a + x = a ' + x; (s)