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Proof step 5 from proof sub-same (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: num-i2n-ls-02 seq: 0 ≤ i ⊦ (* m) < i→n(i) ↔ n→i(m) < i subst: [i, m] → [i, i→n(j) + m] flags: rewrite, current

Additional information

Used simplifier rules: n2i-add: ⊦ n→i(m + n) = n→i(m) + n→i(n);%i2i-pos: ⊦ 0 < i → n→i(i→n(i)) = i;%nat2int-08: ⊦ ¬ n→i(n) < 0;%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: i, i→n(j) + m