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Proof step 5 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 6:
(bv ⊃ P0;P1)(H) ⊦ (bv ⊃ ⌜true⌝ → P0;⌜true⌝ → P1)(H)

Rule argument

Additional information

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