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Proof step 1 from proof short-mod-add-02 (conclusion)

⊦ 0 < s2i(sho) + s2i(sho0) → (s2i(sho) + s2i(sho0)) % 65536 = s2i(sho) + s2i(sho0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 < s2i(sho) + s2i(sho0) ⊦ (s2i(sho) + s2i(sho0)) % 65536 = s2i(sho) + s2i(sho0)

Rule argument

Additional information

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