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Proof step 8 from proof labsp (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: j + 1 abort: F discard: F computed: F

Additional information

Used simplifier rules: rax-04: ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* 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%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used terms: j + 1