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

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: grz: ⊦ 0 < i → (0 < i ^ j ↔ i = 1 ∨ 0 ≤ j);%grz: ⊦ 0 < i → (0 < i ^ j ↔ i = 1 ∨ 0 ≤ j);%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs