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Proof step 3 from proof imp-ite (previous step)

⊦ ((⌜bv0⌝ → (bv ⊃ P0;P1)))(H) ↔ (bv0 → bv ⊃ ⌜bv0⌝ → P0;⌜true⌝ → P1)(H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ bv0 → (bv ⊃ P0;P1)(H) ↔ (bv0 → bv ⊃ ⌜bv0⌝ → P0;⌜true⌝ → P1)(H)

Rule argument

Additional information

Used simplifier rules: imp: ⊦ ((P0 → P1))(H) ↔ P0(H) → P1(H);%prop: ⊦ ⌜bv⌝ (H) ↔ bv;