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Proof step 18 from proof orderedle-all (previous step)

(a0 ' + y)[m0] < a, ordered≤(a0 ' + y), m0 ≤ # y, ∀ m, n. m ≤ n ∧ n < 1 + # y → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦ a0 < a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
n ≠ 0 ⊦ (a ' + x)[(* n)] = x[n -1]
Proof step 20:
y[m0 -1] < a, ordered≤(a0 ' + y), m0 ≤ # y, ∀ m, n. m ≤ n ∧ n < 1 + # y → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦ m0 = 0, a0 < a

Rule argument

spec: list-del name: get-numrec seq: n ≠ 0 ⊦ (a ' + x)[(* n)] = x[n -1] subst: [n, x, a] → [m0, y, a0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s)% ⊦ (a ' + x)[0] = a; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: m0, y, a0