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Proof step 4 from proof imp-ite (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 5:
bv0 → (bv ⊃ P0;P1)(H) ⊦ (bv0 → bv ⊃ ⌜bv0⌝ → P0;⌜true⌝ → P1)(H)
Proof step 7:
bv0, (bv0 → bv ⊃ ⌜bv0⌝ → P0;⌜true⌝ → P1)(H) ⊦ (bv ⊃ P0;P1)(H)

Rule argument

pos: right 1

Additional information