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

uint2bits(j) = boolvar ' + bits, 0 < j, bits.first, boolvar ∈ uint2bits(j) ⊦ true ∈ uint2bits(j)

User applied rule insert elim lemma

And closed the goal!

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0 subst: [bits, boolvar, bits0] → [bits, boolvar0, bits0] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: uint2bits-zero-01: 0 ≤ i ⊦ uint2bits(i) = false ' ↔ i = 0;% ⊦ ¬ boolvar ∈ bits → bits ≠ bits0 + boolvar ' + bits1;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ bits + [] = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ i < j → ¬ j < i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j ∧ i = i0 → i0 < j; (forward)% ⊦ i < j ∧ j = i0 → i < i0; (forward)%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits, boolvar0, bits0