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Proof step 15 from proof orderedle-all (previous step)

(a0 ' + y)[m0] < (a ' + a0 ' + y)[m], ordered≤(a0 ' + y), m ≤ m0 + 1, m ≤ 1 + # 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 16:
n ≠ 0 ⊦ (a ' + x)[(* n)] = x[n -1]
Proof step 17:
m = 0, (a0 ' + y)[m0] < (a ' + a0 ' + y)[m], ordered≤(a0 ' + y), m0 ≤ # y, m ≤ 1 + # y, m ≤ m0 + 1, ∀ m, n. m ≤ n ∧ n < 1 + # y → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦ a0 < a
Proof step 25:
(a0 ' + y)[m0] < (a0 ' + y)[m -1], ordered≤(a0 ' + y), m0 ≤ # y, m ≤ 1 + # y, m ≤ m0 + 1, ∀ m, n. m ≤ n ∧ n < 1 + # y → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦ m = 0, a0 < a

Rule argument

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

Additional information

Used terms: m, a0 ' + y, a