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Proof step 10 from proof and-011x-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
bitsand(false ' + bits0, true ') = false ' + bits1, ∀ bits, bits2, bits1. # bits < # bits0 → 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(# bits0) = 1 + bits2uint(bits1), false ∈ bits0

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)%j ≠ 0 ⊦ 0 % j = 0;% ⊦ 2 ^ n→i(m) ≠ 0;% ⊦ bits2uint([]) = 0;% ⊦ bitsinvert([]) = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;