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Proof step 15 from proof wand-flatten (previous step)

P0(H2), P1(H1), H ⊥ H1, H ⊥ H2, H2 ⊥ H1, ∀ H1. H1 ⊥ H2 ∪ H ∧ P1(H1) → P2(H1 ∪ H2 ∪ H) ⊦ P2((H2 ∪ H1) ∪ H)

User applied rule all left

And obtained the following new goals

Proof step 16:
P0(H2), P1(H1), H2 ⊥ H1, H ⊥ H2, H ⊥ H1, H1 ⊥ H2 ∪ H ∧ P1(H1) → P2(H1 ∪ H2 ∪ H) ⊦ P2((H2 ∪ H1) ∪ H)

Rule argument

pos: left 6 terms: H1 abort: F discard: T computed: F

Additional information

Used terms: H1