[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 169 from proof bytes2int-int2bytes-len-eq (previous step)

256 * b2i(by0) + band(b2i(by), 255) = i + 256, 0 ≤ b2i(by0), ~129 < i, i < 0, ∀ bytes0. # bytes0 < #((by0 + []) + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ 256 * (b2i(by0) - 1) + band(b2i(by), 255) = i, by = i2b(0), by0 ' + by ' = i2b(0) ' + i2b(i) '

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [b2i(by0), 1, i0] rotate: T spec: int-abs inst: name: elim-sub

Additional information

Used simplifier rules: ⊦ band(i, j) = band(j, i);% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i * (i0 + 1) + j = i + j0 ↔ i * i0 + j = j0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: b2i(by0), 1, i0