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Proof step 7 from proof abs-pred-01 (previous step)

i→n(~ i) = m + 1, i < 0 ⊦ i = 0, i = ~1, i→n(i + 1) = m

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: abs-neg: ⊦ i→n(~ i) = i→n(i);%abs-neg: ⊦ i→n(~ i) = i→n(i);%rax-02: ⊦ i + 1 < 0 → i→n(i + 1) = i→n(i) -1;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ n -1 = n - 1; (s)% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)% ⊦ m + n - m = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); (s) Used terms: i + 1