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Proof step 34 from proof sgn-ex-01 (previous step)

mklist(false, # bits0) = bits0, false ∈ bits0 ⊦ true ' + bits0 = mklist(true, 1) + false ' + bits0.rest ∧ sgn(mklist(true, 1) + false ' + bits0.rest) = true ' + false ' + bits0.rest ∧ ¬(true ↔ false) ∧ 1 ≠ 0, ∃ boolvar2, n0, boolvar3, bits1. true ' + bits0 = mklist(boolvar2, n0) + boolvar3 ' + bits1 ∧ sgn(mklist(boolvar2, n0) + boolvar3 ' + bits1) = boolvar2 ' + boolvar3 ' + bits1 ∧ ¬(boolvar2 ↔ boolvar3) ∧ n0 ≠ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
mklist(false, # bits0) = bits0, false ∈ bits0 ⊦ bits0 = false ' + bits0.rest, ∃ boolvar2, n0, boolvar3, bits1. true ' + bits0 = mklist(boolvar2, n0) + boolvar3 ' + bits1 ∧ sgn(mklist(boolvar2, n0) + boolvar3 ' + bits1) = boolvar2 ' + boolvar3 ' + bits1 ∧ ¬(boolvar2 ↔ boolvar3) ∧ n0 ≠ 0

Rule argument

Additional information

Used simplifier rules: sgn-base-01: ⊦ sgn(true ' + false ' + bits) = true ' + false ' + bits;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ mklist(boolvar, 0) = [];%n ≠ 0 ⊦ mklist(true, (* n)) = true ' + mklist(true, n -1);