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Proof step 10 from proof int2bits-len-neg (previous step)

∃ bits. uint2bits(~ j) = false ' + true ' + bits ∧ bits ≠ [], j < 2 ^ i, j < 0, 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 ⊦ # sgn(bitsinvert(uint2bits(~ j)) +1) = i→n(i + 2), true ∈ uint2bits(~ j).rest.rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
uint2bits(~ j) = false ' + true ' + bits, false ∈ uint2bits(~ j), j + 1 < 2 ^ i, j < i, 0 < 2 ^ i, 0 ≤ i, 0 ≤ 2 ^ i * 2 + j, j + 2 ^ i < 0, j + 2 ^ i < i, j < ~1, j < 0, j < 2 ^ i ⊦ bits = [], # sgn(bitsinvert(uint2bits(~ j)) +1) = i→n(i + 2), true ∈ uint2bits(~ j).rest.rest

Rule argument

Additional information

Used simplifier rules: uint2bits-onein-01: ⊦ j < 0 → true ∈ uint2bits(~ j);% ⊦ bits = bits0 + boolvar ' + bits1 → boolvar ∈ bits; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;