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Proof step 6 from proof length-to (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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);% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)