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Proof step 9 from proof ex3-shiftw (previous step)

oP(o)(H0), H0 ⊥ H, ∀ H1. H1 ⊥ H ∧ (∃ o. oP(o)(H1)) → P(H1 ∪ H) ⊦ P(H0 ∪ H)

User applied rule all left

And obtained the following new goals

Proof step 10:
oP(o)(H0), H0 ⊥ H, H0 ⊥ H ∧ (∃ o. oP(o)(H0)) → P(H0 ∪ H), ∀ H1. H1 ⊥ H ∧ (∃ o. oP(o)(H1)) → P(H1 ∪ H) ⊦ P(H0 ∪ H)

Rule argument

pos: left 3 terms: H0 abort: F discard: F computed: T

Additional information

Used terms: H0