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Proof step 4 from proof num-n2i-ls (previous step)

n < i→n(i), i ≤ n→i(n), 0 < i ⊦ i = 0, n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
0 ≤ i ⊦ i→n(i) ≤ (* m) ↔ i ≤ n→i(m)

Rule argument

spec: inst: name: num-i2n-le seq: 0 ≤ i ⊦ i→n(i) ≤ (* m) ↔ i ≤ n→i(m) subst: [m, i] → [n, i] flags: rewrite, current

Additional information

Used simplifier rules: lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: n, i