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Proof step 2 from proof les (previous step)

i < j ⊦ j ≤ i + 1 ↔ j = i + 1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;%numle-01: ⊦ i + (* j) ≤ i0 + (* j0) ↔ i ≤ i0 + (j0 - j);%succ-02: ⊦ i < j + 1 ↔ i ≤ j;%lels-03: ⊦ i < j → (j ≤ i ↔ false);% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)% ⊦ ¬ i < i; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ 0 + i = i; (s)