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Proof step 1 from proof key-size (conclusion) (back to tree)

∀ a. a ∈ st ↔ a ∈ st0 ⊦ # st = # st0

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof key-size (previous step) (back to tree)

# st ≠ # st0, ∀ a. a ∈ st ↔ a ∈ st0 ⊦

User applied rule structural induction

And obtained the following new goals

Rule argument

indvar: st

Additional information

Used simplifier rules: gen: ⊦ ;

Proof step 3 from proof key-size (previous step) (back to tree)

# ∅ ≠ # st0, ∀ a. a ∈ ∅ ↔ a ∈ st0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: Size-empty: ⊦ # ∅ = 0; empty: ⊦ # st = 0 ↔ st = ∅; In-empty: ⊦ ¬ a ∈ ∅;

Proof step 4 from proof key-size (previous step) (back to tree)

st0 ≠ ∅, ∀ a. ¬ a ∈ st0 ⊦

User applied rule structural induction

And obtained the following new goals

Rule argument

indvar: st0

Additional information

Used simplifier rules: gen: ⊦ ;

Proof step 5 from proof key-size (previous step) (back to tree)

st[a := d] ≠ ∅, ∀ a0. ¬ a0 ∈ st[a := d] ⊦ st ≠ ∅ ∧ (∀ a. ¬ a ∈ st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 6 from proof key-size (previous step) (back to tree)

st[a := d] ≠ ∅, ∀ a0. ¬ a0 ∈ st[a := d] ⊦ st ≠ ∅ ∧ ¬ a0 ∈ st

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ st[b := d] ↔ a = b ∨ a ∈ st; Used terms: a

Proof step 7 from proof key-size (previous step) (back to tree)

# st1[a := d] ≠ # st0, ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0, ∀ st0. ¬(# st1 ≠ # st0 ∧ (∀ a. a ∈ st1 ↔ a ∈ st0)) ⊦

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: a ∈ st1

Additional information

Used formulas: a ∈ st1

Proof step 8 from proof key-size (previous step) (back to tree)

# st1[a := d] ≠ # st0, a ∈ st1, ∀ st0. ¬(# st1 ≠ # st0 ∧ (∀ a. a ∈ st1 ↔ a ∈ st0)), ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: size-in: ⊦ a ∈ st → # st[a := d] = # st;

Proof step 9 from proof key-size (previous step) (back to tree)

# st1 ≠ # st0, a ∈ st1, ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0, ∀ st. # st1 = # st ∨ ¬(∀ a. a ∈ st1 ↔ a ∈ st) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 3 terms: a abort: F discard: F computed: T

Additional information

Used terms: a

Proof step 10 from proof key-size (previous step) (back to tree)

a ∈ st1[a := d] ↔ a ∈ st0, # st1 ≠ # st0, a ∈ st1, ∀ st. # st1 = # st ∨ ¬(∀ a. a ∈ st1 ↔ a ∈ st), ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ st[b := d] ↔ a = b ∨ a ∈ st;

Proof step 11 from proof key-size (previous step) (back to tree)

# st1 ≠ # st0, a ∈ st1, a ∈ st1[a := d], a ∈ st0, ∀ st. # st1 = # st ∨ ¬(∀ a. a ∈ st1 ↔ a ∈ st), ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:In-insert, seq: ⊦ a ∈ st[b := d] ↔ a = b ∨ a ∈ st, subst:[d, b, st, a] → [d, a, st1, a], <all paths>

Additional information

Used terms: d, a, st1, a

Proof step 12 from proof key-size (previous step) (back to tree)

⊦ a ∈ st[b := d] ↔ a = b ∨ a ∈ st

Lemma In-insert

Proof step 13 from proof key-size (previous step) (back to tree)

# st1 ≠ # st0, a ∈ st0, a ∈ st1, ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0, ∀ st. # st1 = # st ∨ ¬(∀ a. a ∈ st1 ↔ a ∈ st), a = a ∨ a ∈ st1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 14 from proof key-size (previous step) (back to tree)

# st1 ≠ # st0, a ∈ st1, a ∈ st0, ∀ st. # st1 = # st ∨ ¬(∀ a. a ∈ st1 ↔ a ∈ st), ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0 ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 4 terms: st0 abort: F discard: F computed: T

Additional information

Used terms: st0

Proof step 15 from proof key-size (previous step) (back to tree)

# st1 ≠ # st0, a ∈ st0, a ∈ st1, ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0, ∀ st. # st1 = # st ∨ ¬(∀ a. a ∈ st1 ↔ a ∈ st), # st1 = # st0 ∨ ¬(∀ a. a ∈ st1 ↔ a ∈ st0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 16 from proof key-size (previous step) (back to tree)

# st1 ≠ # st0, ¬(a0 ∈ st1 ↔ a0 ∈ st0), a ∈ st1, a ∈ st0, ∀ st. # st1 = # st ∨ ¬(∀ a. a ∈ st1 ↔ a ∈ st), ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0 ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 6 terms: a0 abort: F discard: F computed: T

Additional information

Used terms: a0

Proof step 17 from proof key-size (previous step) (back to tree)

a0 ∈ st1[a := d] ↔ a0 ∈ st0, # st1 ≠ # st0, ¬(a0 ∈ st1 ↔ a0 ∈ st0), a ∈ st0, a ∈ st1, ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0, ∀ st. # st1 = # st ∨ ¬(∀ a. a ∈ st1 ↔ a ∈ st) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:In-insert, seq: ⊦ a ∈ st[b := d] ↔ a = b ∨ a ∈ st, subst:[d, b, st, a] → [d, a, st1, a0], <all paths>

Additional information

Used terms: d, a, st1, a0

Proof step 18 from proof key-size (previous step) (back to tree)

⊦ a ∈ st[b := d] ↔ a = b ∨ a ∈ st

Lemma In-insert

Proof step 19 from proof key-size (previous step) (back to tree)

# st1[a := d] ≠ # st0, ¬ a ∈ st1, ∀ st0. ¬(# st1 ≠ # st0 ∧ (∀ a. a ∈ st1 ↔ a ∈ st0)), ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: Size-insert: ⊦ ¬ a ∈ st → # st[a := d] = # st + 1;

Proof step 20 from proof key-size (previous step) (back to tree)

# st1 + 1 ≠ # st0, ¬ a ∈ st1, ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0, ∀ st. # st1 = # st ∨ ¬(∀ a. a ∈ st1 ↔ a ∈ st) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 3 terms: a abort: F discard: F computed: T

Additional information

Used terms: a

Proof step 21 from proof key-size (previous step) (back to tree)

a ∈ st1[a := d] ↔ a ∈ st0, # st1 + 1 ≠ # st0, ¬ a ∈ st1, ∀ st. # st1 = # st ∨ ¬(∀ a. a ∈ st1 ↔ a ∈ st), ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ st[b := d] ↔ a = b ∨ a ∈ st;

Proof step 22 from proof key-size (previous step) (back to tree)

# st1 + 1 ≠ # st0, a ∈ st0, ¬ a ∈ st1, ∀ st. # st1 = # st ∨ ¬(∀ a. a ∈ st1 ↔ a ∈ st), ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0 ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 4 terms: st0 -- a abort: F discard: F computed: F

Additional information

Used terms: st0 -- a

Proof step 23 from proof key-size (previous step) (back to tree)

# st1 + 1 ≠ # st0, a ∈ st0, ¬ a ∈ st1, ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0, ∀ st. # st1 = # st ∨ ¬(∀ a. a ∈ st1 ↔ a ∈ st), # st1 = #(st0 -- a) ∨ ¬(∀ a0. a0 ∈ st1 ↔ a0 ∈ st0 -- a) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: size: ⊦ a ∈ st → #(st -- a) = (# st)-1; ⊦ n -1 = n - 1; (s)

Proof step 24 from proof key-size (previous step) (back to tree)

# st1 + 1 ≠ # st0, a ∈ st0, ¬ a ∈ st1, ∀ st. # st1 = # st ∨ ¬(∀ a. a ∈ st1 ↔ a ∈ st), ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0, # st1 = # st0 - 1 ∨ ¬(a0 ∈ st1 ↔ a0 ∈ st0 -- a) ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: nat, name:elim-pred-c, seq:n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1, subst:[n, m] → [# st0, n0]

Additional information

Used simplifier rules: empty: ⊦ # st = 0 ↔ st = ∅; Size-empty: ⊦ # ∅ = 0; In-empty: ⊦ ¬ a ∈ ∅; ⊦ m + 1 ≠ 0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: # st0, m

Proof step 25 from proof key-size (previous step) (back to tree)

# st0 = n0 + 1, # st0 ≠ 0, # st1 + 1 ≠ # st0, a ∈ st0, ¬ a ∈ st1, ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0, ∀ st. # st1 = # st ∨ ¬(∀ a. a ∈ st1 ↔ a ∈ st), # st1 = n0 ∨ ¬(a0 ∈ st1 ↔ a0 ∈ st0 -- a) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: empty: ⊦ # st = 0 ↔ st = ∅; c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 26 from proof key-size (previous step) (back to tree)

# st0 = n0 + 1, # st1 ≠ n0, ¬(a0 ∈ st1 ↔ a0 ∈ st0 -- a), st0 ≠ ∅, a ∈ st0, ¬ a ∈ st1, ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0, ∀ st. # st1 = # st ∨ ¬(∀ a. a ∈ st1 ↔ a ∈ st) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 7 terms: a0 abort: F discard: F computed: F

Additional information

Used terms: a0

Proof step 27 from proof key-size (previous step) (back to tree)

# st0 = n0 + 1, a0 ∈ st1[a := d] ↔ a0 ∈ st0, # st1 ≠ n0, ¬(a0 ∈ st1 ↔ a0 ∈ st0 -- a), st0 ≠ ∅, a ∈ st0, ¬ a ∈ st1, ∀ st. # st1 = # st ∨ ¬(∀ a. a ∈ st1 ↔ a ∈ st), ∀ a0. a0 ∈ st1[a := d] ↔ a0 ∈ st0 ⊦

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 4

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ st[b := d] ↔ a = b ∨ a ∈ st; Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; In-insert: ⊦ a ∈ st[b := d] ↔ a = b ∨ a ∈ st; Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st;

Proof Tree for key-size (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