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Proof step 1 from proof abs-succ (conclusion)

⊦ i→n(i) < i→n(i + 1) ↔ 0 ≤ i

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: rax-04: ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;%abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;%abs-pred-01: ⊦ i < 0 ∧ i ≠ ~1 → i→n(i + 1) = i→n(i) -1;%num-i2n-le-03: i < 0 ⊦ (* m) ≤ i→n(i) ↔ n→i(m) ≤ ~ i;% ⊦ ¬ m + n < m; (s)%n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n; (s)% ⊦ 0 < ~ i ↔ i < 0; (s)% ⊦ 1 ≤ i ↔ 0 < i; (s)% ⊦ n -1 = n - 1; (s)% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ i ≤ j ↔ ¬ j < i; (ss)%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)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2