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Proof step 4 from proof int2int-rest-01 (previous step)

∃ bits. int2bits(j) = true ' + false ' + bits, j < ~1, j < 0 ⊦ bits2int(int2bits(j).rest) = j + 2 ^ (n→i(# int2bits(j)) - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
int2bits(j) = true ' + false ' + bits, j < 0, j < ~1 ⊦ bits2int(int2bits(j).rest) = j + 2 ^ (n→i(# int2bits(j)) - 1), j = 0

Rule argument

Additional information

Used simplifier rules: int2bits-truein: ⊦ true ∈ int2bits(j) ↔ j ≠ 0;% ⊦ bits = bits0 + boolvar ' + bits1 → boolvar ∈ bits;% ⊦ bits = bits0 + boolvar ' + bits1 → boolvar ∈ bits; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)