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Proof step 1 from proof mod-succ (conclusion)

⊦ 1 < i → (i + 1) % i = 1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: mod-add: ⊦ 0 < i ∧ 0 ≤ j → (i + j) % i = j % i;%one-03: ⊦ 1 < j → 1 % j = 1;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1