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Proof step 3 from proof get-from-01 (previous step)

∀ m, m0. m + m0 < # x0 → (x0 from m)[m0] = x0[m + m0] ∨ x0 = [], m + m0 < #(a + x0) ⊦ (a + x0 from m)[m0] = (a + x0)[m + m0], a + x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m + m0 ≤ # x0, ∀ m, m0. m + m0 < # x0 → (x0 from m)[m0] = x0[m + m0] ∨ x0 = [] ⊦ (a ' + x0 from m)[m0] = (a ' + x0)[m + m0]

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ a + x = a ' + x; from specification list