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Proof step 6 from proof short-mod (previous step)

~32768 ≤ s2i(sho) ∧ s2i(sho) ≤ 32767, 65536 ≤ abs(s2i(sho)) ⊦ s2i(sho) % 65536 = s2i(sho), s2i(sho) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
~32768 ≤ s2i(sho), s2i(sho) ≤ 32767, 65536 ≤ abs(s2i(sho)) ⊦ s2i(sho) % 65536 = s2i(sho), s2i(sho) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)