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Proof step 7 from proof labsp (previous step)

0 ≤ i0, 0 ≤ j, ∀ j0. 0 ≤ j0 → (i→n(i0) < i→n(j0) ↔ i0 < j0) ⊦ i→n(i0) + 1 < i→n(j) ↔ i0 + 1 < j

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: j - 1 abort: F discard: T computed: F

Additional information

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