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Proof step 7 from proof to-from-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n0 < # x0, ∀ x0, a. n0 ≤ # x0 ∧ (∀ n. n < # x0 → (x0 to n) from n = [] ∨ x0 = []) → (a ' + x0 to n0) from n0 = [], ∀ n. n < # x0 → (x0 to n) from n = [] ⊦ (x0 to n0) from n0 = [], x0 = []

Rule argument

Additional information

Used simplifier rules: to-succ-cons-l-lem: ⊦ a ' + x to n + 1 = a + (x to n);%from-succ-cons-l-lem: ⊦ a ' + x from n + 1 = x from n;%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%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