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Proof step 4 from proof abs-succ-01 (previous step)

i→n(i) ≤ i→n(i + 1), i < 0 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: labs-01 seq: ⊦ i→n(i) ≤ i→n(j) ↔ abs(i) ≤ abs(j) subst: [j, i] → [i + 1, i] flags: rewrite, current

Additional information

Used simplifier rules: abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%abs-05: ⊦ i < 0 → abs(i + 1) = abs(i) - 1; from specification int-abs%subsub: ⊦ ~ i - j = ~(i + j); from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%negle-13: ⊦ ~ i ≤ j ↔ 0 ≤ i + j; from specification int-abs%neg-15: ⊦ i + ~ j = i - j; from specification int-abs%sub-add-09: ⊦ i - j + i = ~ j; from specification int-abs Used terms: i + 1, i