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Proof step 11 from proof ex-p (previous step)

P(H0), xP(x)(H1), H1 ⊥ H0 ⊦ ∃ H3, H2. H3 ⊥ H2 ∧ H1 ∪ H0 = H3 ∪ H2 ∧ (∃ x. xP(x)(H3)) ∧ P(H2)

User applied rule exists right

And obtained the following new goals

Proof step 12:
P(H0), xP(x)(H1), H1 ⊥ H0 ⊦ H1 ⊥ H0 ∧ H1 ∪ H0 = H1 ∪ H0 ∧ (∃ x. xP(x)(H1)) ∧ P(H0)

Rule argument

pos: right 1 terms: H1, H0 abort: F discard: T computed: T

Additional information

Used terms: H1, H0