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Proof step 8 from proof sgn-ex-01 (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 9:
sgn(mklist(boolvar0, n) + boolvar1 ' + bits) = boolvar0 ' + boolvar1 ' + bits ⊦ n = 0, boolvar0 ↔ boolvar1, ¬ false ∈ mklist(boolvar0, n) ∧ boolvar1 ∧ ¬ false ∈ bits, ¬ true ∈ mklist(boolvar0, n) ∧ ¬ boolvar1 ∧ ¬ true ∈ bits, boolvar ' + mklist(boolvar0, n) + boolvar1 ' + bits = mklist(boolvar0, n + 1) + boolvar1 ' + bits ∧ sgn(mklist(boolvar0, n + 1) + boolvar1 ' + bits) = boolvar0 ' + boolvar1 ' + bits ∧ ¬(boolvar0 ↔ boolvar1) ∧ n + 1 ≠ 0, ∃ boolvar2, n0, boolvar3, bits1. boolvar ' + mklist(boolvar0, n) + boolvar1 ' + bits = mklist(boolvar2, n0) + boolvar3 ' + bits1 ∧ sgn(mklist(boolvar2, n0) + boolvar3 ' + bits1) = boolvar2 ' + boolvar3 ' + bits1 ∧ ¬(boolvar2 ↔ boolvar3) ∧ n0 ≠ 0

Rule argument

pos: right 5 terms: boolvar0, n + 1, boolvar1, bits abort: F discard: F computed: F

Additional information

Used terms: boolvar0, n + 1, boolvar1, bits