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Proof step 18 from proof stability (previous step)

P(H0), P1(H3), H3 ⊥ H2, ∃ H3, H1. H3 ⊥ H1 ∧ H0 = H3 ∪ H1 ∧ P0(H3) ∧ {H2} (H1) ⊦ ((P1 * (P0 -⊗ P)))(H3 ∪ H2), P(H3 ∪ H2), ∃ H. H ⊥ H2 ∧ P0(H) ∧ P(H ∪ H2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
P(H1 ∪ H), P0(H1), P1(H3), H1 ⊥ H, H3 ⊥ H ⊦ ((P1 * (P0 -⊗ P)))(H3 ∪ H), P(H3 ∪ H), ∃ H0. H0 ⊥ H ∧ P0(H0) ∧ P(H0 ∪ H)

Rule argument

Additional information

Used simplifier rules: this: ⊦ {H} (H0) ↔ H = H0;