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Proof step 18 from proof and-first (previous step)

bitsand(boolvar ' + bits1, boolvar0 ' + bits) = boolvar1 ' + bits0, ∀ bits2, boolvar, bits1, boolvar0, boolvar1, bits0. # bits2 < # bits → bitsand(boolvar ' + bits1, boolvar0 ' + bits2) = boolvar1 ' + bits0 → (boolvar1 ↔ boolvar ∧ boolvar0) ⊦ boolvar1 ↔ boolvar ∧ boolvar0, boolvar0 ' + bits = true ', boolvar0 ' + bits = false ', bits1 ≠ [] ∧ bits ≠ []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ bits0 ≠ [] ∧ bits = true ' → bitsand(bits, bits0) = bits0
Proof step 20:
¬(boolvar0 ' + bits ≠ [] ∧ boolvar ' + bits1 = true '), bitsand(boolvar ' + bits1, boolvar0 ' + bits) = boolvar1 ' + bits0, ∀ bits2, boolvar, bits1, boolvar0, boolvar1, bits0. # bits2 < # bits → bitsand(boolvar ' + bits1, boolvar0 ' + bits2) = boolvar1 ' + bits0 → (boolvar1 ↔ boolvar ∧ boolvar0) ⊦ boolvar1 ↔ boolvar ∧ boolvar0, boolvar0 ' + bits = true ', boolvar0 ' + bits = false ', bits1 ≠ [] ∧ bits ≠ []

Rule argument

name: and-rec-03 rotate: F subst: [bits0, bits] → [boolvar0 ' + bits, boolvar ' + bits1]

Additional information

Used simplifier rules: ⊦ boolvar ' = boolvar0 ' + bits ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;%⊦ boolvar ' + bits = boolvar0 ' + bits0 ↔ (boolvar ↔ boolvar0) ∧ bits = bits0;% ⊦ bits = bits + bits0 ↔ bits0 = [];% ⊦ boolvar ' + bits ≠ []; Used terms: boolvar0 ' + bits, boolvar ' + bits1