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Proof step 4 from proof labsp (previous step)

~1 ≤ i0, 0 ≤ j, ∀ j0. 0 ≤ j0 ∧ 0 ≤ i0 → (i→n(i0) < i→n(j0) ↔ i0 < j0) ⊦ i→n(i0 + 1) < i→n(j) ↔ abs(i0 + 1) < j

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: Int2nat-pos seq: ⊦ 0 < i → i→n(i) = i→n(i - 1) + 1 subst: [i] → [i0 + 1] flags: <all paths>

Additional information

Used simplifier rules: abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i < 0 → abs(i + 1) = abs(i) - 1; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ~ i - j = ~(i + j); (s)% ⊦ ~ i < j ↔ 0 < j + i; (s)% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); (s)% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)%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% ⊦ 0 < m ↔ m ≠ 0; (s)%i < 0 ⊦ (* i) + j = j - ~ i; (s)% ⊦ i - j < i0 ↔ i < j + i0; (s)% ⊦ 0 ≤ i → (i < 1 ↔ i = 0); (s) Used terms: i0 + 1