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Proof step 1 from proof abs-pred (conclusion)

⊦ i→n(i - 1) < i→n(i) ↔ 0 < i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i→n(i - 1) < i→n(i), i < 0 ⊦ i = 0

Rule argument

Additional information

Used simplifier rules: sub: j ≤ i, 0 ≤ j ⊦ i→n(i - j) = i→n(i) - i→n(j);%num-i2n-le-02: 0 ≤ i ⊦ (* m) ≤ i→n(i) ↔ n→i(m) ≤ i;%abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)% ⊦ i < j → (j ≤ i ↔ false); (as)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → ¬ j < i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ 1 ≤ i ↔ 0 < i; (s)