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Proof step 13 from proof uint2bits-expos0 (previous step)

uint2bits(i) = boolvar + bits0 ∧ uint2bits(i) ≠ [], uint2bits(i) = false ' + bits1, false ∈ uint2bits(i), 0 < i, bits0.first ⊦ uint2bits(i) = false ' + true ' + bits1.rest, ∃ bits. uint2bits(i) = false ' + true ' + bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
uint2bits(i) = false ' + bits0, bits0.first, 0 < i, false ∈ uint2bits(i) ⊦ uint2bits(i) = false ' + true ' + bits0.rest, ∃ bits. uint2bits(i) = false ' + true ' + bits

Rule argument

Additional information

Used simplifier rules: uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];%⊦ bits = boolvar ' + bits0 ∧ bits = boolvar0 ' + bits1 → (boolvar ↔ boolvar0) ∧ bits0 = bits1; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → ¬ j < i;% ⊦ boolvar + bits = boolvar ' + bits;