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Proof step 8 from proof imp-ite (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

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