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Proof step 31 from proof int2bits-len-neg (previous step)

true ∈ uint2bits(~ j).rest.rest, j < 2 ^ i, j < ~1, j + 2 ^ i < i, j + 2 ^ i < 0, 0 ≤ 2 ^ i * 2 + j, 0 ≤ i, 0 < 2 ^ i, j < i, j + 1 < 2 ^ i, j < 0 ⊦ # bitsinvert(uint2bits(~ j)) +1 = i→n(i + 2)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 32:
⊦ false ∈ bits → # bits +1 = # bits
Proof step 33:
false ∈ bitsinvert(uint2bits(~ j)), true ∈ uint2bits(~ j).rest.rest, j < 2 ^ i, j < ~1, j + 2 ^ i < i, j + 2 ^ i < 0, 0 ≤ 2 ^ i * 2 + j, 0 ≤ i, 0 < 2 ^ i, j < i, j + 1 < 2 ^ i, j < 0 ⊦ # bitsinvert(uint2bits(~ j)) = i→n(i + 2)

Rule argument

seq: ⊦ false ∈ bits → # bits +1 = # bits subst: [bits] → [bitsinvert(uint2bits(~ j))] rotate: F spec: bitopsaux inst: name: succ-len

Additional information

Used simplifier rules: uint2bits-onein-01: ⊦ j < 0 → true ∈ uint2bits(~ j);% ⊦ ¬ false ∈ bits → bits +1 = true + bitsinvert(bits);% ⊦ bitsinvert(bitsinvert(bits)) = bits;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ false ∈ bitsinvert(bits) ↔ true ∈ bits; Used terms: bitsinvert(uint2bits(~ j))