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Proof step 7 from proof notone (previous step)

j = i0 +1, i ^ i0 < i, i ^ i0 ≤ 0, i ^ i0 < 1, 2 ≤ i, 1 < i, 0 < j, i ^ i0 < j, i ^ i0 + 1 < i, ∀ j0, i. i→n(j0) < i→n(j) → true → 1 < i ∧ 0 < j0 → ¬ i ^ j0 ≤ 1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
i ^ i0 < i, i ^ i0 ≤ i0, 1 < i, 2 ≤ i, i ^ i0 < 1, i ^ i0 ≤ 0, 0 ≤ i0, i ^ i0 + 1 < i, ∀ j0, i. i→n(j0) < i→n(i0 +1) → true → 1 < i ∧ 0 < j0 → ¬ i ^ j0 ≤ 1 ⊦

Rule argument

Additional information

Used simplifier rules: forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2