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Proof step 7 from proof rmpos-get (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
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 = [], (a ' + (x0 -1l m1))[n] = (a ' + x0)[n]

Rule argument

Additional information

Used simplifier rules: delpos-rec: ⊦ (a ' + x) -1l (n + 1) = a + (x -1l n);%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%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%c: ⊦ a + x = a ' + x; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2