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Proof step 7 from proof bitsand-fillsgn (previous step)

∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(fillsgn(n, bits), bits0)) = fillsgn(n, bitsand(bits, bits0)), bits0 = [], bits = [], # bits ≠ 1 ∧ # bits0 ≠ 1

User applied rule cut formula

And obtained the following new goals

Proof step 8:
# bits ≠ 1, ∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(fillsgn(n, bits), bits0)) = fillsgn(n, bitsand(bits, bits0)), bits0 = [], bits = [], # bits ≠ 1 ∧ # bits0 ≠ 1
Proof step 17:
∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ # bits ≠ 1, fillsgn(n, bitsand(fillsgn(n, bits), bits0)) = fillsgn(n, bitsand(bits, bits0)), bits0 = [], bits = [], # bits ≠ 1 ∧ # bits0 ≠ 1

Rule argument

fma: # bits ≠ 1

Additional information

Used formulas: # bits ≠ 1