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Proof step 3 from proof uint2bits-succ (previous step)

0 ≤ j, ∀ j0. i→n(j0) < i→n(j) → 0 ≤ j0 → uint2bits(j0 + 1) = sgn(false ' + uint2bits(j0) +1) ⊦ uint2bits(j + 1) = sgn(false ' + uint2bits(j) +1)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: uint2bits-pos-num: 1 < i ⊦ uint2bits((* i)) = uint2bits(i / 2) + (i % 2 = 1);%uint2bits-one: ⊦ uint2bits(1) = false + true;%uint2bits-zero: ⊦ uint2bits(0) = false ';% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + true ')+1 = bits +1 + false;% ⊦ false ' +1 = true ';% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ sgn(false ' + true ' + bits) = false ' + true ' + bits;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ sgn(false ' + true ') = false ' + true '; Used terms: j