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Proof step 2 from proof sgn-ex-01 (previous step)

true ∈ bits, false ∈ bits ⊦ ∃ boolvar, n, boolvar0, bits0. bits = mklist(boolvar, n) + boolvar0 ' + bits0 ∧ sgn(mklist(boolvar, n) + boolvar0 ' + bits0) = boolvar ' + boolvar0 ' + bits0 ∧ ¬(boolvar ↔ boolvar0) ∧ n ≠ 0

User applied rule structural induction

And obtained the following new goals

Proof step 3:
true ∈ bits0 ∧ false ∈ bits0 → (∃ boolvar, n, boolvar0, bits1. bits0 = mklist(boolvar, n) + boolvar0 ' + bits1 ∧ sgn(mklist(boolvar, n) + boolvar0 ' + bits1) = boolvar ' + boolvar0 ' + bits1 ∧ ¬(boolvar ↔ boolvar0) ∧ n ≠ 0), true ∈ boolvar + bits0, false ∈ boolvar + bits0 ⊦ ∃ boolvar1, n, boolvar0, bits1. boolvar + bits0 = mklist(boolvar1, n) + boolvar0 ' + bits1 ∧ sgn(mklist(boolvar1, n) + boolvar0 ' + bits1) = boolvar1 ' + boolvar0 ' + bits1 ∧ ¬(boolvar1 ↔ boolvar0) ∧ n ≠ 0

Rule argument

indvar: bits

Additional information

Used simplifier rules: bitlist freely generated by [], +;;% ⊦ ¬ boolvar ∈ [];