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Proof step 5 from proof length-to (previous step)

∀ n0. n0 < # x0 ∧ (∀ x. n0 ≤ # x → #(x to n0) = n0) → #(x0 to n0 + 1) = n0 + 1 ∨ x0 = [], n0 < #(a + x0), ∀ x. n0 ≤ # x → #(x to n0) = n0 ⊦ #(a + x0 to n0 + 1) = n0 + 1, a + x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n0 ≤ # x0, ∀ n0. n0 < # x0 ∧ (∀ x. n0 ≤ # x → #(x to n0) = n0) → #(x0 to n0 + 1) = n0 + 1 ∨ x0 = [], ∀ x. n0 ≤ # x → #(x to n0) = n0 ⊦ #(x0 to n0) = n0

Rule argument

Additional information

Used simplifier rules: to-succ-cons-l-lem: ⊦ a ' + x to n + 1 = a + (x to n);%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%e-01: ⊦ a ' + x ≠ []; from specification list%eq-01: ⊦ m + n = m + n0 ↔ n = n0; 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%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