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Proof step 2 from proof imp-ite (
previous step
)
⊦ (⌜bv0⌝ → (bv ⊃ P0;P1)) = (bv0 → bv ⊃ ⌜bv0⌝ → P0;⌜true⌝ → P1)
User applied rule expand right
And obtained the following new goals
Proof step 3
:
⊦ ((⌜bv0⌝ → (bv ⊃ P0;P1)))(H) ↔ (bv0 → bv ⊃ ⌜bv0⌝ → P0;⌜true⌝ → P1)(H)
Rule argument
pos: right 1
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