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Proof step 2 from proof abs-succ-01 (previous step)

⊦ i→n(i + 1) < i→n(i) ↔ i < 0

User applied rule case distinction

And obtained the following new goals

Proof step 3:
i < 0 ⊦ i→n(i + 1) < i→n(i)

Rule argument

pos: right 1

Additional information

Used simplifier rules: rax-04: ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;%lels: ⊦ ¬ i < j ↔ j ≤ i; 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%l-02: ⊦ ¬ m + 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