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Proof step 8 from proof ls-int-ls-nat (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: intnat inst: name: num-n2i-ls-04 seq: 0 ≤ i ⊦ n→i(n) ≤ (* i) ↔ n ≤ i→n(i) subst: [i, n] → [i + n→i(n), m] 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%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat%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%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2 Used terms: i + n→i(n), m