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Proof step 28 from proof and-011x-neg (previous step)

bitsand(false ' + bits3, true ' + bits0) = false ' + bits4, ∀ 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(bits4 + boolvar ') + bits2uint(bitsinvert(bits0) + (¬ boolvar)') % 2 ^ (n→i(# bits3) + 1), false ∈ bits3

User applied rule apply induction

And obtained the following new goals

Proof step 29:
2 ^ n→i(# bits3) = 1 + bits2uint(bitsinvert(bits0)) % 2 ^ n→i(# bits3) + bits2uint(bits4) ∨ false ∈ bits3, bitsand(false ' + bits3, true ' + bits0) = false ' + bits4, ∀ 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(bits4 + boolvar ') + bits2uint(bitsinvert(bits0) + (¬ boolvar)') % 2 ^ (n→i(# bits3) + 1), false ∈ bits3

Rule argument

subst: [bits, bits2, bits1] → [bits3, bits0, bits4]

Additional information

Used simplifier rules: ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ ¬ n < n;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: bits3, bits0, bits4