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Proof step 7 from proof uint2bits-expos1 (previous step)

uint2bits(j) = false ' + true ', 1 < j, 0 < j, 0 ≤ j, false ∈ uint2bits(j) ⊦ ∃ bits. uint2bits(j) = false ' + true ' + bits ∧ bits ≠ []

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

name: uint2bits-rec-01 rotate: F subst: [i] → [j]

Additional information

Used simplifier rules: uint2bits-one: ⊦ uint2bits(1) = false + true;%uint2bits-zero: ⊦ uint2bits(0) = false ';%uint2bits-zero-01: 0 ≤ i ⊦ uint2bits(i) = false ' ↔ i = 0;% ⊦ 0 < i → (i / 2 = 0 ↔ i = 1);% ⊦ 0 ≤ i / 2 ↔ ~1 ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i;%⊦ bits + boolvar ' = bits0 + boolvar0 ' ↔ bits = bits0 ∧ (boolvar ↔ boolvar0);% ⊦ bits + boolvar = bits + boolvar ';% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ boolvar ' = boolvar0 ' + bits ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ boolvar ' ≠ []; Used terms: j