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Proof step 39 from proof and-011x-neg (previous step)

bitsand(false ' + bits3, true ' + bits0) = false ' + bits4, 2 ^ n→i(# bits3) = 1 + bits2uint(bitsinvert(bits0)) % 2 ^ n→i(# bits3) + bits2uint(bits4), ∀ bits, bits2, bits1. # bits < #(bits3 + true ') → bitsand(false ' + bits, true ' + bits2) = false ' + bits1 → 2 ^ n→i(# bits) = 1 + bits2uint(bitsinvert(bits2)) % 2 ^ n→i(# bits) + bits2uint(bits1) ∨ false ∈ bits ⊦ 2 ^ (n→i(# bits3) + 1) = 2 * (bits2uint(bits4) + 1) + (bits2uint(bitsinvert(bits0)) * 2) % 2 ^ (n→i(# bits3) + 1), false ∈ bits3

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j subst: [i, j] → [n→i(# bits3), 2] rotate: F spec: int-pot inst: name: succ

Additional information

Used simplifier rules: ⊦ 2 ^ n→i(m) ≠ 0;%0 ≠ j, i0 ≠ 0 ⊦ (i0 * i) % (i0 * j) = i % j * i0;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i * j + i * j0 = i * (j + j0);% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ 0 < i → (i * j = i * j0 ↔ j = j0);% ⊦ 0 ≤ n→i(n); Used terms: n→i(# bits3), 2