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Proof step 6 from proof sgn-ex-01 (previous step)

true ∈ bits0, false ∈ bits0 → bits0 = mklist(boolvar0, n) + boolvar1 ' + bits ∧ sgn(mklist(boolvar0, n) + boolvar1 ' + bits) = boolvar0 ' + boolvar1 ' + bits ∧ ¬(boolvar0 ↔ boolvar1) ∧ n ≠ 0 ⊦ boolvar ∧ ¬ false ∈ bits0, ∃ boolvar2, n0, boolvar3, bits1. boolvar ' + bits0 = mklist(boolvar2, n0) + boolvar3 ' + bits1 ∧ sgn(mklist(boolvar2, n0) + boolvar3 ' + bits1) = boolvar2 ' + boolvar3 ' + bits1 ∧ ¬(boolvar2 ↔ boolvar3) ∧ n0 ≠ 0

User applied rule cut formula

And obtained the following new goals

Proof step 7:
false ∈ bits0, true ∈ bits0, false ∈ bits0 → bits0 = mklist(boolvar0, n) + boolvar1 ' + bits ∧ sgn(mklist(boolvar0, n) + boolvar1 ' + bits) = boolvar0 ' + boolvar1 ' + bits ∧ ¬(boolvar0 ↔ boolvar1) ∧ n ≠ 0 ⊦ boolvar ∧ ¬ false ∈ bits0, ∃ boolvar2, n0, boolvar3, bits1. boolvar ' + bits0 = mklist(boolvar2, n0) + boolvar3 ' + bits1 ∧ sgn(mklist(boolvar2, n0) + boolvar3 ' + bits1) = boolvar2 ' + boolvar3 ' + bits1 ∧ ¬(boolvar2 ↔ boolvar3) ∧ n0 ≠ 0
Proof step 22:
true ∈ bits0, false ∈ bits0 → bits0 = mklist(boolvar0, n) + boolvar1 ' + bits ∧ sgn(mklist(boolvar0, n) + boolvar1 ' + bits) = boolvar0 ' + boolvar1 ' + bits ∧ ¬(boolvar0 ↔ boolvar1) ∧ n ≠ 0 ⊦ false ∈ bits0, boolvar ∧ ¬ false ∈ bits0, ∃ boolvar2, n0, boolvar3, bits1. boolvar ' + bits0 = mklist(boolvar2, n0) + boolvar3 ' + bits1 ∧ sgn(mklist(boolvar2, n0) + boolvar3 ' + bits1) = boolvar2 ' + boolvar3 ' + bits1 ∧ ¬(boolvar2 ↔ boolvar3) ∧ n0 ≠ 0

Rule argument

fma: false ∈ bits0

Additional information

Used formulas: false ∈ bits0