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Proof step 6 from proof labsp (previous step)

0 ≤ j, 0 ≤ i0, ∀ 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;%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%l-num-06: n0 < n ⊦ ¬ m + (* n) < (* n0); from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%zero-gs: ⊦ 0 ≤ i ∧ 0 ≤ j → ¬ i + j < 0; from specification int-abs%ls: n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n; from specification nat%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs Used terms: j - 1