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

false ' + true ' = uint2bits(j), 0 ≤ j, false ∈ uint2bits(j), true ∈ uint2bits(j), ∀ i0, j. i→n(i0) < i→n(1) → uint2bits(i0) = uint2bits(j) ∧ 0 ≤ i0 ∧ 0 ≤ j → i0 = j ⊦ 1 = j

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