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Proof step 2 from proof succ-02 (previous step)

⊦ i < j + 1 ↔ ¬ j < i

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: numls-01: ⊦ i + (* j) < i0 + (* j0) ↔ i < i0 + (j0 - j);%lf-08: ⊦ ¬ i < j ∧ ¬ i0 < i → ¬ i0 < j; (forward) from specification int-basic1%add-less: ⊦ 0 < i → i1 < i1 + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%irref: ⊦ ¬ i < i; from specification int-basic1