Proof step 1 from proof fo-in-propFOs (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(FO(t) ∈ propFOs(t0, l, bel0) ↔ FO(l) ∈ bel0 ∧ t = t0) ⊦

Rule argument

Additional information

Proof step 2 from proof fo-in-propFOs (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 3:

  ¬(FO(t) ∈ propFOs(t0, l, be + bel) ↔ FO(l) ∈ be + bel ∧ t = t0), ∀ t, t0, l. FO(t) ∈ propFOs(t0, l, bel) ↔ FO(l) ∈ bel ∧ t = t0 ⊦

Rule argument

Additional information

Used simplifier rules: propFOs-base: ⊦ propFOs(t, l, []) = []; List('a) freely generated by [], +;; ⊦ ¬ a ∈ []; (s)

Proof step 3 from proof fo-in-propFOs (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 4:

  ¬(FO(t) ∈ propFOs(t0, l, be ') ∨ FO(t) ∈ propFOs(t0, l, bel) ↔ (FO(l) = be ∨ FO(l) ∈ bel) ∧ t = t0), ∀ t, t0, l. FO(t) ∈ propFOs(t0, l, bel) ↔ FO(l) ∈ bel ∧ t = t0 ⊦

Rule argument

Additional information

Used simplifier rules: propFOs-append: ⊦ propFOs(t, l, bel0 + bel1) = propFOs(t, l, bel0) + propFOs(t, l, bel1); a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ∈ a0 ' + x ↔ a = a0 ∨ a ∈ x; (s) ⊦ a + x = a ' + x; (s) ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)

Proof step 4 from proof fo-in-propFOs (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 5:

  be = SF, ¬(FO(t) ∈ propFOs(t0, l, be ') ∨ FO(t) ∈ propFOs(t0, l, bel) ↔ (FO(l) = be ∨ FO(l) ∈ bel) ∧ t = t0), ∀ t, t0, l. FO(t) ∈ propFOs(t0, l, bel) ↔ FO(l) ∈ bel ∧ t = t0 ⊦

• Proof step 7:

  be = W(l0, v), ¬(FO(t) ∈ propFOs(t0, l, be ') ∨ FO(t) ∈ propFOs(t0, l, bel) ↔ (FO(l) = be ∨ FO(l) ∈ bel) ∧ t = t0), ∀ t, t0, l. FO(t) ∈ propFOs(t0, l, bel) ↔ FO(l) ∈ bel ∧ t = t0 ⊦

• Proof step 9:

  be = FO(l0), ¬(FO(t) ∈ propFOs(t0, l, be ') ∨ FO(t) ∈ propFOs(t0, l, bel) ↔ (FO(l) = be ∨ FO(l) ∈ bel) ∧ t = t0), ∀ t, t0, l. FO(t) ∈ propFOs(t0, l, bel) ↔ FO(l) ∈ bel ∧ t = t0 ⊦

Rule argument

Additional information

Used simplifier rules: Bevent freely generated by SF, W, FO;;

Proof step 5 from proof fo-in-propFOs (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  ¬(FO(t) ∈ propFOs(t0, l, bel) ↔ FO(l) ∈ bel ∧ t = t0), ∀ t, t0, l. FO(t) ∈ propFOs(t0, l, bel) ↔ FO(l) ∈ bel ∧ t = t0 ⊦

Rule argument

Additional information

Used simplifier rules: propFOs-one-sf: ⊦ propFOs(t, l, SF ') = []; ⊦ FO(l) ≠ SF; (s) ⊦ ¬ a ∈ []; (s)

Proof step 6 from proof fo-in-propFOs (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: t, t0, l

Proof step 7 from proof fo-in-propFOs (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  ¬(FO(t) ∈ propFOs(t0, l, bel) ↔ FO(l) ∈ bel ∧ t = t0), ∀ t, t0, l. FO(t) ∈ propFOs(t0, l, bel) ↔ FO(l) ∈ bel ∧ t = t0 ⊦

Rule argument

Additional information

Used simplifier rules: propFOs-one-w: ⊦ propFOs(t, l, W(l0, v) ') = []; ⊦ W(l, v) ≠ FO(l0); (s) ⊦ ¬ a ∈ []; (s)

Proof step 8 from proof fo-in-propFOs (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: t, t0, l

Proof step 9 from proof fo-in-propFOs (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  ¬(FO(t) ∈ propFOs(t0, l, FO(l0) ') ∨ FO(t) ∈ propFOs(t0, l, bel) ↔ (l = l0 ∨ FO(l) ∈ bel) ∧ t = t0), ∀ t, t0, l. FO(t) ∈ propFOs(t0, l, bel) ↔ FO(l) ∈ bel ∧ t = t0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ FO(l) = FO(l0) ↔ l = l0; (s)

Proof step 10 from proof fo-in-propFOs (previous step) (back to tree)

Heuristic weak cut applied rule cut formula

And obtained the following new goals

• Proof step 11:

  l0 = l, ¬(FO(t) ∈ propFOs(t0, l, FO(l0) ') ∨ FO(t) ∈ propFOs(t0, l, bel) ↔ (l = l0 ∨ FO(l) ∈ bel) ∧ t = t0), ∀ t, t0, l. FO(t) ∈ propFOs(t0, l, bel) ↔ FO(l) ∈ bel ∧ t = t0 ⊦

• Proof step 13:

  ¬(FO(t) ∈ propFOs(t0, l, FO(l0) ') ∨ FO(t) ∈ propFOs(t0, l, bel) ↔ (l = l0 ∨ FO(l) ∈ bel) ∧ t = t0), l0 ≠ l, ∀ t, t0, l. FO(t) ∈ propFOs(t0, l, bel) ↔ FO(l) ∈ bel ∧ t = t0 ⊦

Rule argument

Additional information

Used formulas: l0 = l

Proof step 11 from proof fo-in-propFOs (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  t ≠ t0, FO(t) ∈ propFOs(t0, l, bel), ∀ t, t0, l. FO(t) ∈ propFOs(t0, l, bel) ↔ FO(l) ∈ bel ∧ t = t0 ⊦

Rule argument

Additional information

Used simplifier rules: propFOs-one-fo-same: ⊦ propFOs(t, l, FO(l) ') = FO(t) '; ⊦ a ∈ a0 ' ↔ a = a0; (s) ⊦ FO(tid) = FO(tid0) ↔ tid = tid0; (s)

Proof step 12 from proof fo-in-propFOs (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: t, t0, l

Proof step 13 from proof fo-in-propFOs (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  ¬(FO(t) ∈ propFOs(t0, l, bel) ↔ FO(l) ∈ bel ∧ t = t0), l0 ≠ l, ∀ t, t0, l. FO(t) ∈ propFOs(t0, l, bel) ↔ FO(l) ∈ bel ∧ t = t0 ⊦

Rule argument

Additional information

Used simplifier rules: propFOs-one-fo-other: ⊦ l0 ≠ l → propFOs(t, l, FO(l0) ') = []; ⊦ ¬ a ∈ []; (s)

Proof step 14 from proof fo-in-propFOs (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: t, t0, l