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Proof step 6 from proof le-pos (previous step)

¬ 0 < j, 2 ≤ i, 0 ≤ j, 1 < i, ∀ j1, i, j0. i→n(j1) < i→n(j) → 2 ≤ i ∧ 0 ≤ j1 ∧ 1 < i → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ i ^ j ≤ i ^ j0 ↔ j ≤ j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 < i, 0 ≤ i, 2 ≤ i, ∀ j1, i, j0. i→n(j1) < i→n(0) → 2 ≤ i ∧ 0 ≤ j1 ∧ 1 < i → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ 0 < i ^ j0 ↔ 0 ≤ j0

Rule argument

Additional information

Used simplifier rules: Pot-zero: ⊦ j ^ 0 = 1;%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%forward-le: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%numlsledel-01: i0 ≠ 0, i0 < i ⊦ (* i) ≤ j → (* i0) < j; from specification int-abs