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Proof step 9 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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ 1 < # bits ∧ 1 < # bits0 → bitsand(bits, bits0) = bitsand(bits.butlast, bits0.butlast) + (bits.last ∧ bits0.last)
Proof step 11:
¬(1 < #(boolvar ' + bits1) ∧ 1 < #(boolvar0 ' + bits)), 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
Proof step 25:
1 < #(boolvar ' + bits1) ∧ 1 < #(boolvar0 ' + bits), bitsand((boolvar ' + bits1).butlast, (boolvar0 ' + bits).butlast) + ((boolvar ' + bits1).last ∧ (boolvar0 ' + bits).last) = boolvar1 ' + bits0, ∀ bits2, boolvar, bits1, boolvar0, boolvar1, bits0. # bits2 < # bits → bitsand(boolvar ' + bits1, boolvar0 ' + bits2) = boolvar1 ' + bits0 → (boolvar1 ↔ boolvar ∧ boolvar0) ⊦ boolvar1 ↔ boolvar ∧ boolvar0

Rule argument

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

Additional information

Used terms: boolvar0 ' + bits, boolvar ' + bits1