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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

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;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%ls-01: n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; from specification nat%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs