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Proof step 1 from proof short-succ-pos (conclusion)

⊦ 0 ≤ s2i(sho) ∧ s2i(sho) < s2i(shoi) → s2i(i2s(s2i(sho) + 1)) = s2i(sho) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
s2i(sho) < s2i(shoi), 0 ≤ s2i(sho), 0 < s2i(shoi) ⊦ s2i(i2s(s2i(sho) + 1)) = s2i(sho) + 1

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)