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Proof step 3 from proof int2bits-one-bit (previous step)

int2bits(i) = boolvar ' ⊦ i = ~1 ∧ boolvar, i = 0 ∧ ¬ boolvar

User applied rule cut formula

And obtained the following new goals

Proof step 4:
i < 0, int2bits(i) = boolvar ' ⊦ i = ~1 ∧ boolvar, i = 0 ∧ ¬ boolvar

Rule argument

fma: i < 0

Additional information

Used simplifier rules: int2bits-mone-02: ⊦ int2bits(~1) = true ';%int2bits-true: ⊦ int2bits(j) = true ' ↔ j = ~1;%int2bits-eq: ⊦ int2bits(j) = false ' ↔ j = 0;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) Used formulas: i < 0