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Proof step 7 from proof bits2uint-succ (previous step)

∀ bits1. # bits1 < #(bits0 + boolvar ') → true → bits1 = [] ∨ bits2uint(bits1 +1) = bits2uint(bits1) + 1 ⊦ bits2uint((bits0 + boolvar ')+1) = bits2uint(bits0 + boolvar ') + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ boolvar → (bits + boolvar ')+1 = bits +1 + false
Proof step 9:
boolvar, ∀ bits1. # bits1 < #(bits0 + boolvar ') → true → bits1 = [] ∨ bits2uint(bits1 +1) = bits2uint(bits1) + 1 ⊦ bits2uint(bits0 +1 + false) = bits2uint(bits0 + boolvar ') + 1

Rule argument

seq: ⊦ boolvar → (bits + boolvar ')+1 = bits +1 + false subst: [boolvar, bits] → [boolvar, bits0] rotate: F spec: bitopsaux inst: name: succ-rec-01

Additional information

Used simplifier rules: bits2uint-one: ⊦ bits2uint(bits + true ') = bits2uint(bits) * 2 + 1;%bits2uint-zero: ⊦ bits2uint(bits + false ') = bits2uint(bits) * 2;% ⊦ i * j = j * i;% ⊦ (bits + false ')+1 = bits + true;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ i + j = i + i0 ↔ j = i0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used terms: boolvar, bits0