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Proof step 8 from proof and-011x-neg (previous step)

¬(1 < #(false ' + bits0) ∧ 1 < #(true ' + bits2)), 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 ⊦ 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 9:
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 ⊦ 2 ^ n→i(# bits0) = 1 + bits2uint(bitsinvert(bits2)) % 2 ^ n→i(# bits0) + bits2uint(bits1), false ∈ bits0, bits0 ≠ [] ∧ bits2 ≠ []

Rule argument

Additional information

Used simplifier rules: and-zero-01: ⊦ bitsand(false ', boolvar ' + bits) = false ';% ⊦ ¬ boolvar ∈ [];% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ i % 1 = 0;% ⊦ # [] = 0;% ⊦ boolvar ' = boolvar0 ' + bits ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ bits + [] = bits;% ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 = m + 1 ↔ m = 0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≠ 0;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;