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Proof step 45 from proof and-insgn (previous step)

∀ bits1, bits, boolvar. # bits1 < #(bits2 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ bits2 + boolvar0 ' = (bits2 + boolvar0 ').first ', bits0 + true ' = (bits0 + true ').first ', (boolvar ↔ boolvar0) ∨ boolvar ∈ bits2 ↔ boolvar ∈ bitsand(bits0, bits2) ∨ (boolvar ↔ boolvar0), false ∈ bits0

User applied rule case distinction

And obtained the following new goals

Proof step 46:
(boolvar ↔ boolvar0) ∨ boolvar ∈ bits2, ∀ bits1, bits, boolvar. # bits1 < #(bits2 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ boolvar ∈ bitsand(bits0, bits2) ∨ (boolvar ↔ boolvar0), bits2 + boolvar0 ' = (bits2 + boolvar0 ').first ', bits0 + true ' = (bits0 + true ').first ', false ∈ bits0
Proof step 49:
boolvar ∈ bitsand(bits0, bits2) ∨ (boolvar ↔ boolvar0), ∀ bits1, bits, boolvar. # bits1 < #(bits2 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ (boolvar ↔ boolvar0) ∨ boolvar ∈ bits2, bits2 + boolvar0 ' = (bits2 + boolvar0 ').first ', bits0 + true ' = (bits0 + true ').first ', false ∈ bits0

Rule argument

pos: right 3

Proof step 45 from proof and-insgn (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information