Proof step 1 from proof del-three (conclusion) (back to tree)

⊦ 2 < # s → #(s -- a -- b) ≠ 0

System applied rule normalize

And obtained the following new goals

Proof step 2:
#(s -- a -- b) = 0, 2 < # s ⊦

Rule argument

Additional information

Proof step 2 from proof del-three (previous step) (back to tree)

#(s -- a -- b) = 0, 2 < # s ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
#(s -- a -- b) = 0, 2 < # s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Rule argument

Additional information

Used simplifier rules: Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s; ⊦ m < n → n ≠ 0; (s) ⊦ ¬ a ∈ ∅; (s) ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic size-zero-02: ⊦ # s = 0 → s = ∅; (forward) from specification set-basic

Proof step 3 from proof del-three (previous step) (back to tree)

#(s -- a -- b) = 0, 2 < # s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 4:
#(s -- a -- b) = 0, 2 < # s, b ∈ s -- a, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦
Proof step 12:
#(s -- a -- b) = 0, 2 < # s, ¬ b ∈ s -- a, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Rule argument

fma: b ∈ s -- a

Additional information

Used formulas: b ∈ s -- a

Proof step 4 from proof del-three (previous step) (back to tree)

#(s -- a -- b) = 0, 2 < # s, b ∈ s -- a, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
#(s -- a) - 1 = 0, b ≠ a, 2 < # s, b ∈ s, a0 ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Rule argument

Additional information

Used simplifier rules: del-size: ⊦ a ∈ s → #(s -- a) = # s - 1; Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s; ⊦ ¬ a ∈ ∅; (s) ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s) ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic

Proof step 5 from proof del-three (previous step) (back to tree)

#(s -- a) - 1 = 0, b ≠ a, 2 < # s, b ∈ s, a0 ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 6:
#(s -- a) - 1 = 0, b ≠ a, 2 < # s, a0 ∈ s, b ∈ s, a ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦
Proof step 10:
#(s -- a) - 1 = 0, b ≠ a, 2 < # s, a0 ∈ s, b ∈ s, ¬ a ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Rule argument

fma: a ∈ s

Additional information

Used formulas: a ∈ s

Proof step 6 from proof del-three (previous step) (back to tree)

#(s -- a) - 1 = 0, b ≠ a, 2 < # s, a0 ∈ s, b ∈ s, a ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# s - 1 - 1 = 0, b ≠ a, 2 < # s, a ∈ s, b ∈ s, a0 ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Rule argument

Additional information

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

Proof step 7 from proof del-three (previous step) (back to tree)

# s - 1 - 1 = 0, b ≠ a, 2 < # s, a ∈ s, b ∈ s, a0 ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Proof step 8:
# s = n0 + 1, n0 - 1 = 0, # s ≠ 0, b ≠ a, 2 < # s, a0 ∈ s, b ∈ s, a ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Rule argument

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

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: # s, m

Proof step 8 from proof del-three (previous step) (back to tree)

# s = n0 + 1, n0 - 1 = 0, # s ≠ 0, b ≠ a, 2 < # s, a0 ∈ s, b ∈ s, a ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# s = n0 + 1, n0 - 1 = 0, b ≠ a, n0 ≠ 1, n0 ≠ 0, a0 ∈ s, b ∈ s, a ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s) n0 < n ⊦ (* n) < m0 + (* n0) ↔ n - n0 < m0; (s) ⊦ m + 1 ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m < n → n ≠ 0; (s)

Proof step 9 from proof del-three (previous step) (back to tree)

# s = n0 + 1, n0 - 1 = 0, b ≠ a, n0 ≠ 1, n0 ≠ 0, a0 ∈ s, b ∈ s, a ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Heuristic elimination applied rule apply elim lemma

And closed the goal!

Rule argument

spec: nat, name:elim-pred-c, seq:n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1, subst:[n, m] → [n0, n1]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n0, m

Proof step 10 from proof del-three (previous step) (back to tree)

#(s -- a) - 1 = 0, b ≠ a, 2 < # s, a0 ∈ s, b ∈ s, ¬ a ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# s - 1 = 0, b ≠ a, 2 < # s, b ∈ s, a0 ∈ s, ¬ a ∈ s, ∀ a1. a1 = b ∨ ¬ a1 ∈ s ⊦

Rule argument

Additional information

Used simplifier rules: del-in: ⊦ ¬ a ∈ s → s -- a = s;

Proof step 11 from proof del-three (previous step) (back to tree)

# s - 1 = 0, b ≠ a, 2 < # s, b ∈ s, a0 ∈ s, ¬ a ∈ s, ∀ a1. a1 = b ∨ ¬ a1 ∈ s ⊦

Heuristic elimination applied rule apply elim lemma

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) size-one-02: ⊦ # s = 1 ∧ a ∈ s ∧ b ∈ s → a = b; (forward) from specification set-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: # s, m

Proof step 12 from proof del-three (previous step) (back to tree)

#(s -- a -- b) = 0, 2 < # s, ¬ b ∈ s -- a, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
#(s -- a) = 0, 2 < # s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s ⊦ b ≠ a ∧ b ∈ s

Rule argument

Additional information

Used simplifier rules: del-in: ⊦ ¬ a ∈ s → s -- a = s; Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s; ⊦ ¬ a ∈ ∅; (s) ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s) size-zero-02: ⊦ # s = 0 → s = ∅; (forward) from specification set-basic

Proof step 13 from proof del-three (previous step) (back to tree)

#(s -- a) = 0, 2 < # s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s ⊦ b ≠ a ∧ b ∈ s

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 14:
#(s -- a) = 0, 2 < # s, a ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦ b ≠ a ∧ b ∈ s

Rule argument

fma: a ∈ s

Additional information

Used simplifier rules: del-in: ⊦ ¬ a ∈ s → s -- a = s; ⊦ m < n → n ≠ 0; (s) Used formulas: a ∈ s

Proof step 14 from proof del-three (previous step) (back to tree)

#(s -- a) = 0, 2 < # s, a ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦ b ≠ a ∧ b ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# s - 1 = 0, 2 < # s, a ∈ s, a0 ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦ b ≠ a ∧ b ∈ s

Rule argument

Additional information

Used simplifier rules: del-size: ⊦ a ∈ s → #(s -- a) = # s - 1; ⊦ ¬ a ∈ ∅; (s) ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s) ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic

Proof step 15 from proof del-three (previous step) (back to tree)

# s - 1 = 0, 2 < # s, a ∈ s, a0 ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦ b ≠ a ∧ b ∈ s

Heuristic elimination applied rule apply elim lemma

And closed the goal!

Rule argument

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

Additional information

Proof Tree for del-three (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.