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

i < 0, i < 1, i→n(1 - i) < i→n(i) ⊦

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] → [1 - i] flags: <all paths>

Additional information

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