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Proof step 10 from proof fits-law-div (previous step)

b2i(by1) / i = b2i(i2b(b2i(by1) / i)) ⊦ i = 0, b2i(by1) / i = s2i(i2s(b2i(by1) / i)), i = ~1, ¬ b2i(by1) / i < ~32768 ∧ ¬ 32767 < b2i(by1) / i

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: (b2i(by1) / i)∈byte

Additional information

Used simplifier rules: bib-in: ⊦ i ∈byte → b2i(i2b(i)) = i;%inbyte: ⊦ i ∈byte ↔ ~128 ≤ i ∧ i ≤ 127;%short-max-10: ⊦ 32767 < i → s2i(sho) ≠ i;% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) Used formulas: (b2i(by1) / i)∈byte