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Proof step 7 from proof le-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
0 < i, 0 ≤ i, 2 ≤ i, ∀ j1, i, j0. i→n(j1) < i→n(0) → 1 < i ∧ 2 ≤ i ∧ 0 ≤ j1 → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ 0 < i ^ j0 ↔ 0 ≤ j0

Rule argument

Additional information

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