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Proof step 3 from proof le-int-le-nat (previous step)

0 ≤ i ⊦ i + n→i(n) ≤ n→i(m) ↔ i→n(i) + n ≤ m

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: intnat inst: name: num-n2i-ls-06 seq: 0 ≤ i ⊦ (* i) ≤ n→i(n) ↔ i→n(i) ≤ n subst: [n, i] → [m, i + n→i(n)] flags: rewrite, current

Additional information

Used simplifier rules: nat2int-08: ⊦ ¬ n→i(n) < 0;%add-nonneg-int-to-nat: ⊦ 0 ≤ i → i→n(i + n→i(n)) = i→n(i) + n;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2 Used terms: m, i + n→i(n)