Proof step 1 from proof modified-stays-reachable (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(reachable(r, r0, h[r0, ce]) ↔ true), reachable(r, r0, h) ⊦

Rule argument

Additional information

Proof step 2 from proof modified-stays-reachable (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  reachable(r, r0, h), ¬ reachable(r, r0, h[r0, ce]) ⊦

Rule argument

Additional information

Proof step 3 from proof modified-stays-reachable (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

  ⊦ reachable(r, r0, h) ↔ (∃ x. path(r ' + x, h) ∧ (r ' + x).last = r0)

• Proof step 5:

  reachable(r, r0, h), ∀ x. ¬(path(r ' + x, h[r0, ce]) ∧ (r ' + x).last = r0) ⊦

Rule argument

Additional information

Used terms: h[r0, ce], r0, r

Proof step 4 from proof modified-stays-reachable (previous step) (back to tree)

Lemma reachable-def

Proof step 5 from proof modified-stays-reachable (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  reachable(r, r0, h), ∀ x. ¬ path(r ' + x, h[(r ' + x).last, ce]) ∨ (r ' + x).last ≠ r0 ⊦

Rule argument

Additional information

Proof step 6 from proof modified-stays-reachable (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

  ⊦ reachable(r, r0, h) ↔ (∃ x. path(r ' + x, h) ∧ (r ' + x).last = r0)

• Proof step 8:

  ∀ x. ¬ path(r ' + x, h[(r ' + x).last, ce]) ∨ (r ' + x).last ≠ r0, ∃ x. path(r ' + x, h) ∧ (r ' + x).last = r0 ⊦

Rule argument

Additional information

Used terms: h, r0, r

Proof step 7 from proof modified-stays-reachable (previous step) (back to tree)

Lemma reachable-def

Proof step 8 from proof modified-stays-reachable (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  path(r ' + x, h), ∀ x0. ¬ path(r ' + x0, h[(r ' + x0).last, ce]) ∨ (r ' + x0).last ≠ (r ' + x).last ⊦

Rule argument

Additional information

Proof step 9 from proof modified-stays-reachable (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 10:

  path(r ' + [], h), ∀ x0. ¬ path(r ' + x0, h[(r ' + x0).last, ce]) ∨ (r ' + x0).last ≠ (r ' + []).last ⊦

• Proof step 12:

  path(r ' + (r0 + x0), h), ∀ x1. ¬ path(r ' + x1, h[(r ' + x1).last, ce]) ∨ (r ' + x1).last ≠ (r ' + (r0 + x0)).last, ∀ r, h, ce. ¬(path(r ' + x0, h) ∧ (∀ x1. ¬ path(r ' + x1, h[(r ' + x1).last, ce]) ∨ (r ' + x1).last ≠ (r ' + x0).last)) ⊦

Rule argument

Additional information

Used simplifier rules: reflist freely generated by [], +;;

Proof step 10 from proof modified-stays-reachable (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  r ≠ null, r ∈ h, ∀ x0. ¬ path(r ' + x0, h[(r ' + x0).last, ce]) ∨ (r ' + x0).last ≠ r ⊦

Rule argument

Additional information

Used simplifier rules: path-one: ⊦ path(r ', h) ↔ r ∈ h ∧ r ≠ null; ⊦ a '.last = a; (s) a: ⊦ (x + y) + z = x + y + z; from specification list[reflist] ⊦ x + [] = x; (s)

Proof step 11 from proof modified-stays-reachable (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: path-one: ⊦ path(r ', h) ↔ r ∈ h ∧ r ≠ null; ⊦ a ∈ st[a, d]; (s) ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list[reflist] ⊦ a '.last = a; (s) Used terms: []

Proof step 12 from proof modified-stays-reachable (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  r ≠ null, path(h[r].nxt ' + x0, h), r ∈ h, ∀ x1. ¬ path(r ' + x1, h[(r ' + x1).last, ce]) ∨ (r ' + x1).last ≠ (h[r].nxt ' + x0).last, ∀ r1, h0, ce0. ¬ path(r1 ' + x0, h0) ∨ ¬(∀ x1. ¬ path(r1 ' + x1, h0[(r1 ' + x1).last, ce0]) ∨ (r1 ' + x1).last ≠ (r1 ' + x0).last) ⊦

Rule argument

Additional information

Used simplifier rules: path-cons: ⊦ path(r ' + r0 ' + x, h) ↔ r ∈ h ∧ r ≠ null ∧ h[r].nxt = r0 ∧ path(r0 ' + x, h); a: ⊦ (x + y) + z = x + y + z; from specification list[reflist] ⊦ (b ' + a ' + x).last = (a ' + x).last; (s) ⊦ a + x = a ' + x; (s)

Proof step 13 from proof modified-stays-reachable (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 14:

  r ≠ null, path(h[r].nxt ' + x0, h), r ∈ h, ∀ x1. ¬ path(r ' + x1, h[(r ' + x1).last, ce]) ∨ (r ' + x1).last ≠ (h[r].nxt ' + x0).last, ¬ path(h[r].nxt ' + x0, h) ∨ ¬(∀ x1. ¬ path(h[r].nxt ' + x1, h[(h[r].nxt ' + x1).last, ce]) ∨ (h[r].nxt ' + x1).last ≠ (h[r].nxt ' + x0).last) ⊦

Rule argument

Additional information

Used terms: h[r].nxt, h, ce

Proof step 14 from proof modified-stays-reachable (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  (h[r].nxt ' + x1).last = (h[r].nxt ' + x0).last, r ≠ null, path(h[r].nxt ' + x0, h), path(h[r].nxt ' + x1, h[(h[r].nxt ' + x0).last, ce]), r ∈ h, ∀ x1. ¬ path(r ' + x1, h[(r ' + x1).last, ce]) ∨ (r ' + x1).last ≠ (h[r].nxt ' + x0).last ⊦

Rule argument

Additional information

Proof step 15 from proof modified-stays-reachable (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 16:

  (h[r].nxt ' + x1).last = (h[r].nxt ' + x0).last, r ≠ null, path(h[r].nxt ' + x1, h[(h[r].nxt ' + x0).last, ce]), path(h[r].nxt ' + x0, h), r ∈ h, ∀ x1. ¬ path(r ' + x1, h[(r ' + x1).last, ce]) ∨ (r ' + x1).last ≠ (h[r].nxt ' + x0).last, ¬ path(r ' + (h[r].nxt + x1), h[(r ' + (h[r].nxt + x1)).last, ce]) ∨ (r ' + (h[r].nxt + x1)).last ≠ (h[r].nxt ' + x0).last ⊦

Rule argument

Additional information

Used terms: h[r].nxt + x1

Proof step 16 from proof modified-stays-reachable (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 17:

  (h[r].nxt ' + x1).last = (h[r].nxt ' + x0).last, h[(h[r].nxt ' + x0).last, ce][r].nxt ≠ h[r].nxt, r ≠ null, path(h[r].nxt ' + x1, h[(h[r].nxt ' + x0).last, ce]), path(h[r].nxt ' + x0, h), r ∈ h, r ∈ h[(h[r].nxt ' + x0).last, ce], ∀ x1. ¬ path(r ' + x1, h[(r ' + x1).last, ce]) ∨ (r ' + x1).last ≠ (h[r].nxt ' + x0).last ⊦

Rule argument

Additional information

Used simplifier rules: path-cons: ⊦ path(r ' + r0 ' + x, h) ↔ r ∈ h ∧ r ≠ null ∧ h[r].nxt = r0 ∧ path(r0 ' + x, h); ⊦ ¬ a ∈ st → st[a] = ∅[a]; (s) ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st; (ws) a: ⊦ (x + y) + z = x + y + z; from specification list[reflist] ⊦ (b ' + a ' + x).last = (a ' + x).last; (s) ⊦ a + x = a ' + x; (s)

Proof step 17 from proof modified-stays-reachable (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: path-one: ⊦ path(r ', h) ↔ r ∈ h ∧ r ≠ null; ⊦ a ∈ st[a, d]; (s) ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list[reflist] ⊦ a '.last = a; (s) ⊦ a ≠ b → st[b, d][a] = st[a]; (s) Used terms: []