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Proof step 8 from proof abs-pred (previous step)

i < 0, i→n(1 - i) < i→n(i) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: intnat inst: name: Int2nat-pos seq: ⊦ 0 < i → i→n(i) = i→n(i - 1) + 1 subst: [i] → [1 - i] flags: rewrite, current

Additional information

Used simplifier rules: sub: j ≤ i, 0 ≤ j ⊦ i→n(i - j) = i→n(i) - i→n(j);%num-i2n-le-01: i < 0 ⊦ i→n(i) ≤ (* m) ↔ ~ i ≤ n→i(m);%abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;%abs-neg: ⊦ i→n(~ i) = i→n(i);%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; 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%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%sub-sub: ⊦ i - j - i0 = i - j + i0; from specification int-abs%num-subadd: ⊦ (* i) - j + (* i0) = i - i0 - j; from specification int-abs%neg-06: ⊦ 0 - i = ~ i; from specification int-abs%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%negle-13: ⊦ ~ i ≤ j ↔ 0 ≤ i + j; from specification int-abs%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%mone-08: ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); from specification int-abs%ls-01: n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; from specification nat%two-02: ⊦ m + m = 2 * m; from specification nat-mult%ls2-03: ⊦ 1 < 2 * n ↔ n ≠ 0; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: 1 - i