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Proof step 1 from proof sharing-ass (conclusion) (back to tree)

⊦ hP0 ⊗ hP1 ⊗ hP2 = (hP0 ⊗ hP1) ⊗ hP2

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof sharing-ass (previous step) (back to tree)

hP0 ⊗ hP1 ⊗ hP2 ≠ (hP0 ⊗ hP1) ⊗ hP2 ⊦

User applied rule expand left

And obtained the following new goals

Rule argument

pos: left 1

Additional information

Proof step 3 from proof sharing-ass (previous step) (back to tree)

¬((hP0 ⊗ hP1 ⊗ hP2)(h) ↔ ((hP0 ⊗ hP1) ⊗ hP2)(h)) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:sharing-ass-aux1, seq:⊦ (hP0 ⊗ hP1 ⊗ hP2)(h) ↔ (∃ h0, h1, h2, h21, h20, h10, h210. h = (h1 ∪ h0 ∪ h10) ∪ h2 ∪ h20 ∪ h21 ∪ h210 ∧ h0 ⊥ h1 ∧ h0 ⊥ h2 ∧ h0 ⊥ h21 ∧ h0 ⊥ h20 ∧ h0 ⊥ h10 ∧ h0 ⊥ h210 ∧ h1 ⊥ h2 ∧ h1 ⊥ h21 ∧ h1 ⊥ h20 ∧ h1 ⊥ h10 ∧ h1 ⊥ h210 ∧ h2 ⊥ h21 ∧ h2 ⊥ h20 ∧ h2 ⊥ h10 ∧ h2 ⊥ h210 ∧ h21 ⊥ h20 ∧ h21 ⊥ h10 ∧ h21 ⊥ h210 ∧ h20 ⊥ h10 ∧ h20 ⊥ h210 ∧ h10 ⊥ h210 ∧ hP0(h0 ∪ h10 ∪ h20 ∪ h210) ∧ hP1(h1 ∪ h10 ∪ h21 ∪ h210) ∧ hP2(h2 ∪ h20 ∪ h21 ∪ h210)), subst:[hP0, hP1, hP2, h] → [hP0, hP1, hP2, h]; [A, B] → [A, B], lhs:(hP0 ⊗ hP1 ⊗ hP2)(h), <all paths>

Additional information

Used terms: hP0, hP1, hP2, h

Proof step 4 from proof sharing-ass (previous step) (back to tree)

⊦ (hP0 ⊗ hP1 ⊗ hP2)(h) ↔ (∃ h0, h1, h2, h21, h20, h10, h210. h = (h1 ∪ h0 ∪ h10) ∪ h2 ∪ h20 ∪ h21 ∪ h210 ∧ h0 ⊥ h1 ∧ h0 ⊥ h2 ∧ h0 ⊥ h21 ∧ h0 ⊥ h20 ∧ h0 ⊥ h10 ∧ h0 ⊥ h210 ∧ h1 ⊥ h2 ∧ h1 ⊥ h21 ∧ h1 ⊥ h20 ∧ h1 ⊥ h10 ∧ h1 ⊥ h210 ∧ h2 ⊥ h21 ∧ h2 ⊥ h20 ∧ h2 ⊥ h10 ∧ h2 ⊥ h210 ∧ h21 ⊥ h20 ∧ h21 ⊥ h10 ∧ h21 ⊥ h210 ∧ h20 ⊥ h10 ∧ h20 ⊥ h210 ∧ h10 ⊥ h210 ∧ hP0(h0 ∪ h10 ∪ h20 ∪ h210) ∧ hP1(h1 ∪ h10 ∪ h21 ∪ h210) ∧ hP2(h2 ∪ h20 ∪ h21 ∪ h210))

Lemma sharing-ass-aux1

Proof step 5 from proof sharing-ass (previous step) (back to tree)

⊦ (∃ h0, h1, h2, h21, h20, h10, h210. h = (h1 ∪ h0 ∪ h10) ∪ h2 ∪ h20 ∪ h21 ∪ h210 ∧ h0 ⊥ h1 ∧ h0 ⊥ h2 ∧ h0 ⊥ h21 ∧ h0 ⊥ h20 ∧ h0 ⊥ h10 ∧ h0 ⊥ h210 ∧ h1 ⊥ h2 ∧ h1 ⊥ h21 ∧ h1 ⊥ h20 ∧ h1 ⊥ h10 ∧ h1 ⊥ h210 ∧ h2 ⊥ h21 ∧ h2 ⊥ h20 ∧ h2 ⊥ h10 ∧ h2 ⊥ h210 ∧ h21 ⊥ h20 ∧ h21 ⊥ h10 ∧ h21 ⊥ h210 ∧ h20 ⊥ h10 ∧ h20 ⊥ h210 ∧ h10 ⊥ h210 ∧ hP0(h0 ∪ h10 ∪ h20 ∪ h210) ∧ hP1(h1 ∪ h10 ∪ h21 ∪ h210) ∧ hP2(h2 ∪ h20 ∪ h21 ∪ h210)) ↔ ((hP0 ⊗ hP1) ⊗ hP2)(h)

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:sharing-ass-aux2, seq:⊦ ((hP0 ⊗ hP1) ⊗ hP2)(h) ↔ (∃ h0, h1, h2, h21, h20, h10, h210. h = (h1 ∪ h0 ∪ h10) ∪ h2 ∪ h20 ∪ h21 ∪ h210 ∧ h0 ⊥ h1 ∧ h0 ⊥ h2 ∧ h0 ⊥ h21 ∧ h0 ⊥ h20 ∧ h0 ⊥ h10 ∧ h0 ⊥ h210 ∧ h1 ⊥ h2 ∧ h1 ⊥ h21 ∧ h1 ⊥ h20 ∧ h1 ⊥ h10 ∧ h1 ⊥ h210 ∧ h2 ⊥ h21 ∧ h2 ⊥ h20 ∧ h2 ⊥ h10 ∧ h2 ⊥ h210 ∧ h21 ⊥ h20 ∧ h21 ⊥ h10 ∧ h21 ⊥ h210 ∧ h20 ⊥ h10 ∧ h20 ⊥ h210 ∧ h10 ⊥ h210 ∧ hP0(h0 ∪ h10 ∪ h20 ∪ h210) ∧ hP1(h1 ∪ h10 ∪ h21 ∪ h210) ∧ hP2(h2 ∪ h20 ∪ h21 ∪ h210)), subst:[hP0, hP1, hP2, h] → [hP0, hP1, hP2, h]; [A, B] → [A, B], lhs:((hP0 ⊗ hP1) ⊗ hP2)(h), <all paths>

Additional information

Used terms: hP0, hP1, hP2, h

Proof step 6 from proof sharing-ass (previous step) (back to tree)

⊦ ((hP0 ⊗ hP1) ⊗ hP2)(h) ↔ (∃ h0, h1, h2, h21, h20, h10, h210. h = (h1 ∪ h0 ∪ h10) ∪ h2 ∪ h20 ∪ h21 ∪ h210 ∧ h0 ⊥ h1 ∧ h0 ⊥ h2 ∧ h0 ⊥ h21 ∧ h0 ⊥ h20 ∧ h0 ⊥ h10 ∧ h0 ⊥ h210 ∧ h1 ⊥ h2 ∧ h1 ⊥ h21 ∧ h1 ⊥ h20 ∧ h1 ⊥ h10 ∧ h1 ⊥ h210 ∧ h2 ⊥ h21 ∧ h2 ⊥ h20 ∧ h2 ⊥ h10 ∧ h2 ⊥ h210 ∧ h21 ⊥ h20 ∧ h21 ⊥ h10 ∧ h21 ⊥ h210 ∧ h20 ⊥ h10 ∧ h20 ⊥ h210 ∧ h10 ⊥ h210 ∧ hP0(h0 ∪ h10 ∪ h20 ∪ h210) ∧ hP1(h1 ∪ h10 ∪ h21 ∪ h210) ∧ hP2(h2 ∪ h20 ∪ h21 ∪ h210))

Lemma sharing-ass-aux2

Proof Tree for sharing-ass (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.

the proof tree