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Proof step 10 from proof orderedle-all (previous step)

m ≤ 1 + # y, n ≤ 1 + # y, ordered≤(a0 ' + y), m ≤ n, (a ' + a0 ' + y)[n] < (a ' + a0 ' + y)[m], ∀ 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 11:
n ≠ 0 ⊦ (a ' + x)[(* n)] = x[n -1]
Proof step 12:
n ≠ 0, m ≤ 1 + # y, n ≤ 1 + # y, ordered≤(a0 ' + y), m ≤ n, (a0 ' + y)[n -1] < (a ' + a0 ' + y)[m], ∀ m, n. m ≤ n ∧ n < 1 + # y → ¬ (a0 ' + y)[n] < (a0 ' + y)[m] ⊦ a0 < a

Rule argument

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

Additional information

Used simplifier rules: lez: ⊦ 0 ≤ m; from specification nat%lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat%get-zero-01: ⊦ (a ' + x)[0] = a; from specification list-del%irreflexivity: ⊦ ¬ a < a; from specification oelem%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: n, a0 ' + y, a