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Proof step 3 from proof uint2uint-inv (previous step)

0 < i, ∀ i0. i→n(i0) < i→n(i) → 0 < i0 → 2 ^ n→i(# uint2bits(i0)) = bits2uint(bitsinvert(uint2bits(i0))) + i0 + 1 ⊦ 2 ^ n→i(# uint2bits(i)) = bits2uint(bitsinvert(uint2bits(i))) + i + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

name: uint2bits-rec-01 rotate: F subst: [i] → [i]

Additional information

Used simplifier rules: uint2bits-one: ⊦ uint2bits(1) = false + true;%bits2uint-zero: ⊦ bits2uint(bits + false ') = bits2uint(bits) * 2;%bits2uint-one-03: ⊦ bits2uint(true ') = 1;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ i < j ∧ j = i0 → i < i0; (forward)% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bitsinvert(false ' + bits) = true + bitsinvert(bits);% ⊦ bitsinvert(true ') = false ';% ⊦ boolvar + bits = boolvar ' + bits; Used terms: i