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

⊦ # st0 < # st → ¬ st ⊆ st0

System applied rule normalize

And obtained the following new goals

Proof step 2:
# st0 < # st, st ⊆ st0 ⊦

Rule argument

Additional information

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

# st0 < # st, st ⊆ st0 ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 3:
# st0 < # st1[elem, data], st1[elem, data] ⊆ st0, ∀ st0. ¬(# st0 < # st1 ∧ st1 ⊆ st0) ⊦

Rule argument

indvar: st

Additional information

Used simplifier rules: gen: ⊦ ; subset-empty: ⊦ ∅ ⊆ st; Size-empty: ⊦ # ∅ = 0; ⊦ ¬ n < 0; (s)

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

# st0 < # st1[elem, data], st1[elem, data] ⊆ st0, ∀ st0. ¬(# st0 < # st1 ∧ st1 ⊆ st0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
st1[elem, data] ≠ ∅, # st0 < # st1[elem, data], st1[elem, data] ⊆ st0, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Rule argument

Additional information

Used simplifier rules: empty: ⊦ # st = 0 ↔ st = ∅; f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

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

st1[elem, data] ≠ ∅, # st0 < # st1[elem, data], st1[elem, data] ⊆ st0, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 5:
st1[elem, data] ≠ ∅, # st0 < # st1[elem, data], elem ∈ st1, st1[elem, data] ⊆ st0, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦
Proof step 18:
st1[elem, data] ≠ ∅, # st0 < # st1[elem, data], st1[elem, data] ⊆ st0, ¬ elem ∈ st1, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Rule argument

fma: elem ∈ st1

Additional information

Used formulas: elem ∈ st1

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

st1[elem, data] ≠ ∅, # st0 < # st1[elem, data], elem ∈ st1, st1[elem, data] ⊆ st0, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
st1[elem, data] ≠ ∅, st1 ≠ ∅, # st0 < # st1, elem ∈ st1, st1[elem, data] ⊆ st0, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Rule argument

Additional information

Used simplifier rules: size-in: ⊦ a ∈ st → # st[a, d] = # st; empty: ⊦ # st = 0 ↔ st = ∅; f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

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

st1[elem, data] ≠ ∅, st1 ≠ ∅, # st0 < # st1, elem ∈ st1, st1[elem, data] ⊆ st0, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ st[a] = d ∧ a ∈ st → st[a, d] = st
Proof step 8:
st1[elem, data] ≠ ∅, st1 ≠ ∅, # st0 < # st1, elem ∈ st1, st1[elem, data] ⊆ st0, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦ st1[elem] = data ∧ elem ∈ st1
Proof step 16:
st1[elem] = data, st1 ≠ ∅, # st0 < # st1, elem ∈ st1, st1 ⊆ st0, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Rule argument

spec: <toplevel>, name:ins, seq: ⊦ st[a] = d ∧ a ∈ st → st[a, d] = st, subst:[d, a, st] → [data, elem, st1], <all paths>

Additional information

Used terms: data, elem, st1

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

⊦ st[a] = d ∧ a ∈ st → st[a, d] = st

Lemma ins

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

st1[elem, data] ≠ ∅, st1 ≠ ∅, # st0 < # st1, elem ∈ st1, st1[elem, data] ⊆ st0, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦ st1[elem] = data ∧ elem ∈ st1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
st1[elem, data] ≠ ∅, st1[elem] ≠ data, st1 ≠ ∅, # st0 < # st1, elem ∈ st1, st1[elem, data] ⊆ st0, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Rule argument

Additional information

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

st1[elem, data] ≠ ∅, st1[elem] ≠ data, st1 ≠ ∅, # st0 < # st1, elem ∈ st1, st1[elem, data] ⊆ st0, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ st[a, d] ⊆ st0 ↔ st -- a ⊆ st0 -- a ∧ a ∈ st0 ∧ st0[a] = d
Proof step 11:
st0[elem] = data, st1[elem, data] ≠ ∅, st1[elem] ≠ data, st1 ≠ ∅, # st0 < # st1, elem ∈ st0, elem ∈ st1, st1 -- elem ⊆ st0 -- elem, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Rule argument

spec: <toplevel>, name:subset-put-02, seq: ⊦ st[a, d] ⊆ st0 ↔ st -- a ⊆ st0 -- a ∧ a ∈ st0 ∧ st0[a] = d, subst:[st0, d, a, st] → [st0, data, elem, st1], <all paths>

Additional information

Used terms: st0, data, elem, st1

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

⊦ st[a, d] ⊆ st0 ↔ st -- a ⊆ st0 -- a ∧ a ∈ st0 ∧ st0[a] = d

Lemma subset-put-02

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

st0[elem] = data, st1[elem, data] ≠ ∅, st1[elem] ≠ data, st1 ≠ ∅, # st0 < # st1, elem ∈ st0, elem ∈ st1, st1 -- elem ⊆ st0 -- elem, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
st1[elem, st0[elem]] ≠ ∅, st1[elem] ≠ st0[elem], st1 ≠ ∅, # st0 < # st1, elem ∈ st1, elem ∈ st0, st1 -- elem ⊆ st0 -- elem, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Rule argument

Additional information

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

st1[elem, st0[elem]] ≠ ∅, st1[elem] ≠ st0[elem], st1 ≠ ∅, # st0 < # st1, elem ∈ st1, elem ∈ st0, st1 -- elem ⊆ st0 -- elem, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

User applied rule all left

And obtained the following new goals

Proof step 13:
st1[elem, st0[elem]] ≠ ∅, st1[elem] ≠ st0[elem], st1 ≠ ∅, # st0 < # st1, elem ∈ st0, elem ∈ st1, st1 -- elem ⊆ st0 -- elem, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st, ¬ # st0[elem, st1[elem]] < # st1 ∨ ¬ st1 ⊆ st0[elem, st1[elem]] ⊦

Rule argument

pos: left 8 terms: st0[elem, st1[elem]] abort: F discard: F computed: F

Additional information

Used terms: st0[elem, st1[elem]]

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

st1[elem, st0[elem]] ≠ ∅, st1[elem] ≠ st0[elem], st1 ≠ ∅, # st0 < # st1, elem ∈ st0, elem ∈ st1, st1 -- elem ⊆ st0 -- elem, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st, ¬ # st0[elem, st1[elem]] < # st1 ∨ ¬ st1 ⊆ st0[elem, st1[elem]] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
st1[elem, st0[elem]] ≠ ∅, st1[elem] ≠ st0[elem], st1 ≠ ∅, # st0 < # st1, elem ∈ st1, elem ∈ st0, st1 -- elem ⊆ st0 -- elem, ¬ st1 ⊆ st0[elem, st1[elem]], ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Rule argument

Additional information

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

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

st1[elem, st0[elem]] ≠ ∅, st1[elem] ≠ st0[elem], st1 ≠ ∅, # st0 < # st1, elem ∈ st1, elem ∈ st0, st1 -- elem ⊆ st0 -- elem, ¬ st1 ⊆ st0[elem, st1[elem]], ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ st ⊆ st0[a, d] ↔ st ⊆ st0 ∧ ¬ a ∈ st ∨ a ∈ st ∧ st[a] = d ∧ st -- a ⊆ st0 -- a

Rule argument

spec: <toplevel>, name:subset-put-01, seq:⊦ st ⊆ st0[a, d] ↔ st ⊆ st0 ∧ ¬ a ∈ st ∨ a ∈ st ∧ st[a] = d ∧ st -- a ⊆ st0 -- a, subst:[d, a, st0, st] → [st1[elem], elem, st0, st1], <all paths>

Additional information

Used terms: st1[elem], elem, st0, st1

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

⊦ st ⊆ st0[a, d] ↔ st ⊆ st0 ∧ ¬ a ∈ st ∨ a ∈ st ∧ st[a] = d ∧ st -- a ⊆ st0 -- a

Lemma subset-put-01

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

st1[elem] = data, st1 ≠ ∅, # st0 < # st1, elem ∈ st1, st1 ⊆ st0, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
st1 ≠ ∅, # st0 < # st1, elem ∈ st1, st1 ⊆ st0, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Rule argument

Additional information

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

st1 ≠ ∅, # st0 < # st1, elem ∈ st1, st1 ⊆ st0, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: st0

Additional information

Used terms: st0

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

st1[elem, data] ≠ ∅, # st0 < # st1[elem, data], st1[elem, data] ⊆ st0, ¬ elem ∈ st1, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
st1[elem, data] ≠ ∅, # st0 ≤ # st1, st1[elem, data] ⊆ st0, ¬ elem ∈ st1, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Rule argument

Additional information

Used simplifier rules: Size-insert: ⊦ ¬ a ∈ st → # st[a, d] = # st + 1; ⊦ m < n ↔ ¬ n ≤ m; (ss) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < n + 1 ↔ ¬ n < m; (s)

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

st1[elem, data] ≠ ∅, # st0 ≤ # st1, st1[elem, data] ⊆ st0, ¬ elem ∈ st1, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ st[a, d] ⊆ st0 ↔ st -- a ⊆ st0 -- a ∧ a ∈ st0 ∧ st0[a] = d
Proof step 21:
st0[elem] = data, st1[elem, data] ≠ ∅, elem ∈ st0, # st0 ≤ # st1, st1 -- elem ⊆ st0 -- elem, ¬ elem ∈ st1, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Rule argument

spec: <toplevel>, name:subset-put-02, seq: ⊦ st[a, d] ⊆ st0 ↔ st -- a ⊆ st0 -- a ∧ a ∈ st0 ∧ st0[a] = d, subst:[st0, d, a, st] → [st0, data, elem, st1], <all paths>

Additional information

Used terms: st0, data, elem, st1

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

⊦ st[a, d] ⊆ st0 ↔ st -- a ⊆ st0 -- a ∧ a ∈ st0 ∧ st0[a] = d

Lemma subset-put-02

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

st0[elem] = data, st1[elem, data] ≠ ∅, elem ∈ st0, # st0 ≤ # st1, st1 -- elem ⊆ st0 -- elem, ¬ elem ∈ st1, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
st1[elem, st0[elem]] ≠ ∅, elem ∈ st0, # st0 ≤ # st1, st1 ⊆ st0 -- elem, ¬ elem ∈ st1, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Rule argument

Additional information

Used simplifier rules: notin-del-notin: ⊦ ¬ a ∈ st → st -- a = st;

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

st1[elem, st0[elem]] ≠ ∅, elem ∈ st0, # st0 ≤ # st1, st1 ⊆ st0 -- elem, ¬ elem ∈ st1, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: st0 -- elem

Additional information

Used simplifier rules: size: ⊦ a ∈ st → #(st -- a) = (# st)-1; empty: ⊦ # st = 0 ↔ st = ∅; Del-empty: ⊦ ∅ -- a = ∅; subset-empty-01: ⊦ st ⊆ ∅ ↔ st = ∅; In-empty: ⊦ ¬ a ∈ ∅; ⊦ n -1 = n - 1; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m < 1 ↔ m = 0; (s) n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m ≤ n → (n < m ↔ false); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: st0 -- elem

Proof Tree for subset-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.