[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 82 from proof bitsand-fillsgn (previous step)

∀ n, bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ bits3 = [], bits2 = [], fillsgn(m + 1, bitsand(fillsgn(m, bits2), bits3) + (boolvar ∧ boolvar0)') = fillsgn(m + 1, bitsand(bits2, bits3) + (boolvar ∧ boolvar0)'), fillsgn(m, bits2) = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 83:
⊦ n ≠ 0 ∧ bits ≠ [] → fillsgn(n, bits + boolvar ') = fillsgn(n -1, bits) + boolvar
Proof step 84:
m + 1 ≠ 0 ∧ bitsand(fillsgn(m, bits2), bits3) ≠ [], ∀ n, bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ bits3 = [], bits2 = [], fillsgn((m + 1)-1, bitsand(fillsgn(m, bits2), bits3)) + (boolvar ∧ boolvar0) = fillsgn(m + 1, bitsand(bits2, bits3) + (boolvar ∧ boolvar0)'), fillsgn(m, bits2) = []

Rule argument

seq: ⊦ n ≠ 0 ∧ bits ≠ [] → fillsgn(n, bits + boolvar ') = fillsgn(n -1, bits) + boolvar subst: [boolvar, bits, n] → [boolvar ∧ boolvar0, bitsand(fillsgn(m, bits2), bits3), m + 1] rotate: F spec: bitopsaux inst: name: fill-last

Additional information

Used simplifier rules: and-notempty: ⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) ≠ [];% ⊦ m + 1 ≠ 0;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: boolvar ∧ boolvar0, bitsand(fillsgn(m, bits2), bits3), m + 1