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Proof step 21 from proof and-011x-neg (previous step)

bitsand(false ' + bits3, true ' + bits0) + boolvar ' = false ' + bits1, ∀ bits, bits2, bits1. # bits < #(bits3 + true ') → bitsand(false ' + bits, true ' + bits2) = false ' + bits1 → 2 ^ n→i(# bits) = 1 + bits2uint(bitsinvert(bits2)) % 2 ^ n→i(# bits) + bits2uint(bits1) ∨ false ∈ bits ⊦ 2 ^ (n→i(# bits3) + 1) = 1 + bits2uint(bitsinvert(bits0) + (¬ boolvar)') % 2 ^ (n→i(# bits3) + 1) + bits2uint(bits1), false ∈ bits3, ¬ false ∈ bitsand(false ' + bits3, true ' + bits0) ∧ boolvar, ¬ boolvar ∈ bits1 ∧ boolvar, boolvar ∧ ¬ false ∈ bitsand(false ' + bits3, true ' + bits0)

User applied rule cut formula

And obtained the following new goals

Proof step 22:
bitsand(false ' + bits3, true ' + bits0) + boolvar ' = false ' + bits1, ∀ bits, bits2, bits1. # bits < #(bits3 + true ') → bitsand(false ' + bits, true ' + bits2) = false ' + bits1 → 2 ^ n→i(# bits) = 1 + bits2uint(bitsinvert(bits2)) % 2 ^ n→i(# bits) + bits2uint(bits1) ∨ false ∈ bits ⊦ bits1 = [], 2 ^ (n→i(# bits3) + 1) = 1 + bits2uint(bitsinvert(bits0) + (¬ boolvar)') % 2 ^ (n→i(# bits3) + 1) + bits2uint(bits1), false ∈ bits3, ¬ false ∈ bitsand(false ' + bits3, true ' + bits0) ∧ boolvar, ¬ boolvar ∈ bits1 ∧ boolvar, boolvar ∧ ¬ false ∈ bitsand(false ' + bits3, true ' + bits0)

Rule argument

fma: bits1 = []

Additional information

Used simplifier rules: and-notempty-05: ⊦ bitsand(boolvar0 ' + bits, boolvar ' + bits0) ≠ [];% ⊦ bits + [] = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' = bits + boolvar0 ' ↔ (boolvar ↔ boolvar0) ∧ bits = []; Used formulas: bits1 = []