Proof step 1 from proof lls-fold-tail (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  x ≠ [], (r1 => x.last × r2 * lls(r0, r1, x.butlast) * hP)(h), ¬ (lls(r0, r2, x) * hP)(h) ⊦

Rule argument

Additional information

Proof step 2 from proof lls-fold-tail (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  x ≠ [], (r1 => x.last × r2 * lls(r0, r1, x.butlast) * hP)(h), ¬ (lls(r0, r2, x) * hP)(h) ⊦

Rule argument

Additional information

Proof step 3 from proof lls-fold-tail (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 4:

  a + x0 ≠ [], (r1 => (a + x0).last × r2 * lls(r0, r1, (a + x0).butlast) * hP)(h), ¬ (lls(r0, r2, a + x0) * hP)(h), ∀ r1, r2, r0, hP, h. ¬(x0 ≠ [] ∧ (r1 => x0.last × r2 * lls(r0, r1, x0.butlast) * hP)(h) ∧ ¬ (lls(r0, r2, x0) * hP)(h)) ⊦

Rule argument

Additional information

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

Proof step 4 from proof lls-fold-tail (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  (r1 => (a ' + x0).last × r2 * lls(r0, r1, (a ' + x0).butlast) * hP)(h), x0 = [] ∨ (∀ r, r3, r4, hP0, h0. ¬ (r => x0.last × r3 * lls(r4, r, x0.butlast) * hP0)(h0) ∨ (lls(r4, r3, x0) * hP0)(h0)) ⊦ ∃ r. ((r0 => a × r * lls(r, r2, x0)) * hP)(h)

Rule argument

Additional information

Used simplifier rules: lls-rec: ⊦ lls(r, r1, a ' + x) = _∃(λ r0. r => a × r0 * lls(r0, r1, x)); sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification map-sep sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification map-sep ⊦ (_∃ bP * hP)(h) ↔ (∃ b. (bP(b) * hP)(h)); (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a + x = a ' + x; (s) ⊦ a ' ≠ []; (s) ⊦ x + y = [] ↔ x = [] ∧ y = []; (s)

Proof step 5 from proof lls-fold-tail (previous step) (back to tree)

Heuristic weak cut applied rule cut formula

And obtained the following new goals

• Proof step 6:

  x0 ≠ [], (r1 => (a ' + x0).last × r2 * lls(r0, r1, (a ' + x0).butlast) * hP)(h), x0 = [] ∨ (∀ r, r3, r4, hP0, h0. ¬ (r => x0.last × r3 * lls(r4, r, x0.butlast) * hP0)(h0) ∨ (lls(r4, r3, x0) * hP0)(h0)) ⊦ ∃ r. ((r0 => a × r * lls(r, r2, x0)) * hP)(h)

Rule argument

Additional information

Used simplifier rules: lls-base: ⊦ lls(r0, r1, []) = (emp ∧ ⌜r0 = r1⌝ ); ⊦ a '.last = a; (s) ⊦ x + [] = x; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a '.butlast = []; (s) ⊦ emp * hP = hP; (s) ⊦ (⌜bv⌝ ∧ hP0) * hP1 = (⌜bv⌝ ∧ hP0 * hP1); (s) sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification map-sep sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification map-sep ⊦ (hP0 ∧ hP1)(h) ↔ hP0(h) ∧ hP1(h); (s) ⊦ ⌜bv⌝ (h) ↔ bv; (s) and-ass: ⊦ ((hP0 ∧ hP1) ∧ hP2) = (hP0 ∧ hP1 ∧ hP2); from specification map-sep and-com: ⊦ (hP0 ∧ hP1) = (hP1 ∧ hP0); from specification map-sep Used formulas: x0 = []

Proof step 6 from proof lls-fold-tail (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  x0 ≠ [], ((r0 => a × r * lls(r, r1, x0.butlast)) * r1 => x0.last × r2 * hP)(h), ∀ r, r3, r4, hP0, h0. ¬ (r => x0.last × r3 * lls(r4, r, x0.butlast) * hP0)(h0) ∨ (lls(r4, r3, x0) * hP0)(h0) ⊦ ∃ r. ((r0 => a × r * lls(r, r2, x0)) * hP)(h)

Rule argument

Additional information

Used simplifier rules: lls-rec: ⊦ lls(r, r1, a ' + x) = _∃(λ r0. r => a × r0 * lls(r0, r1, x)); sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification map-sep sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification map-sep ⊦ (_∃ bP * hP)(h) ↔ (∃ b. (bP(b) * hP)(h)); (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ x ≠ [] → (a ' + x).butlast = a ' + x.butlast; (s) ⊦ x ≠ [] → (a ' + x).last = x.last; (s)

Proof step 7 from proof lls-fold-tail (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 8:

  x0 ≠ [], ((r0 => a × r * lls(r, r1, x0.butlast)) * r1 => x0.last × r2 * hP)(h), ¬ (r1 => x0.last × r2 * lls(r, r1, x0.butlast) * hP * r0 => a × r)(h) ∨ (lls(r, r2, x0) * hP * r0 => a × r)(h) ⊦ ∃ r. ((r0 => a × r * lls(r, r2, x0)) * hP)(h)

Rule argument

Additional information

Used terms: r1, r2, r, hP * r0 => a × r, h

Proof step 8 from proof lls-fold-tail (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  x0 ≠ [], (lls(r, r2, x0) * hP * r0 => a × r)(h), ((r0 => a × r * lls(r, r1, x0.butlast)) * r1 => x0.last × r2 * hP)(h) ⊦ ∃ r. ((r0 => a × r * lls(r, r2, x0)) * hP)(h)

Rule argument

Additional information

Used simplifier rules: sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification map-sep sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification map-sep eq-objs: ⊦ (r => o0 * hP0)(h) ∧ (r => o1 * hP1)(h) → o0 = o1; (forward) from specification refmap-maplet

Proof step 9 from proof lls-fold-tail (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification map-sep sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification map-sep Used terms: r