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Proof step 1 from proof imp-ite (conclusion)

⊦ (⌜bv0⌝ → ite(bv, P0, P1)) = ite(bv0 → bv, ⌜bv0⌝ → P0, ⌜bv0⌝ → P1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (⌜bv0⌝ → (bv ⊃ P0;P1)) = (bv0 → bv ⊃ ⌜bv0⌝ → P0;⌜true⌝ → P1)

Rule argument

Additional information

Used simplifier rules: ite: ⊦ ite(bv, P0, P1) = (bv ⊃ P0;P1);