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Proof step 7 from proof bits2uint-not-neg (previous step)

∀ bits1. # bits1 < #(bits0 + boolvar ') → true → (bits2uint(bits1) ≤ 0 ↔ bits2uint(bits1) = 0) ⊦ bits2uint(bits0 + boolvar ') ≤ 0 ↔ bits2uint(bits0 + boolvar ') = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ boolvar → bits2uint(bits + boolvar ') = bits2uint(bits) * 2 + 1
Proof step 9:
¬ boolvar, ∀ bits1. # bits1 < #(bits0 + boolvar ') → true → (bits2uint(bits1) ≤ 0 ↔ bits2uint(bits1) = 0) ⊦ bits2uint(bits0 + boolvar ') ≤ 0 ↔ bits2uint(bits0 + boolvar ') = 0
Proof step 11:
boolvar, ∀ bits1. # bits1 < #(bits0 + boolvar ') → true → (bits2uint(bits1) ≤ 0 ↔ bits2uint(bits1) = 0) ⊦ bits2uint(bits0) * 2 + 1 ≤ 0 ↔ bits2uint(bits0) * 2 + 1 = 0

Rule argument

name: bits2uint-rec-02 rotate: F subst: [boolvar, bits] → [boolvar, bits0]

Additional information

Used terms: boolvar, bits0