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Proof step 22 from proof int2bits-truein-01 (previous step)

uint2bits(j) = false ' + true ' + bits, false ∈ uint2bits(j), 0 < j ⊦ j = 0, true ∈ uint2bits(j).rest, ∃ i. ¬ i < 0 ∧ 2 ^ i + j = 0

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 1 flags: discard

Additional information

Used simplifier rules: ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (boolvar ' + bits).rest = bits;%⊦ ¬ boolvar ∈ boolvar0 ' + bits ↔ ¬(boolvar ↔ boolvar0) ∧ ¬ boolvar ∈ bits;