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Proof step 12 from proof to-to-leq (previous step)

n0 < # x, ∀ m, x. m ≤ n0 ∧ n0 ≤ # x ∧ m ≤ # x → x to n0 to m = x to m ⊦ (x to n0) + x[n0] ' to n0 + 1 = x to n0 + 1, x = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
m < # x ⊦ x to m + 1 = (x to m) + x[m] '
Proof step 14:
n0 < # x, n0 < #((x to n0) + x[n0] '), ∀ m, x. m ≤ n0 ∧ n0 ≤ # x ∧ m ≤ # x → x to n0 to m = x to m ⊦ ((x to n0) + x[n0] ' to n0) + ((x to n0) + x[n0] ')[n0] ' = x to n0 + 1, x = []

Rule argument

spec: <toplevel> name: to-unwind-end-fin seq: m < # x ⊦ x to m + 1 = (x to m) + x[m] ' subst: [m, x] → [n0, (x to n0) + x[n0] '] flags: <all paths>

Additional information

Used simplifier rules: length-to: ¬ # x < n ⊦ #(x to n) = n;% ⊦ #(x + y) = # x + # y; (s)% ⊦ m < n → ¬ n < m; (s)% ⊦ # a ' = 0 +1; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n0, (x to n0) + x[n0] '