Proof step 1 from proof bmin-bin (conclusion) (back to tree)

⊦ s ≠ ∅ ∧ # s ≠ 1 → s.butmin.min ∈ s

System applied rule normalize

And obtained the following new goals

Proof step 2:
# s ≠ 1, s ≠ ∅, ¬ s.butmin.min ∈ s ⊦

Rule argument

Additional information

Proof step 2 from proof bmin-bin (previous step) (back to tree)

# s ≠ 1, s ≠ ∅, ¬ s.butmin.min ∈ s ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# s ≠ 1, # s ≠ 0, s ≠ ∅, s.min ∈ s, s.max ∈ s, ¬ s.max < s.min, ¬ (s -- s.min).min ∈ s ⊦

Rule argument

Additional information

Used simplifier rules: Butmin: ⊦ s ≠ ∅ → s.butmin = s -- s.min; Max-in: ⊦ s ≠ ∅ → s.max ∈ s; Min-in: ⊦ s ≠ ∅ → s.min ∈ s; Min-min: ⊦ a ∈ s → ¬ a < s.min; Max-max: ⊦ a ∈ s → ¬ s.max < a; ⊦ ¬ a < a; (s) ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic

Proof step 3 from proof bmin-bin (previous step) (back to tree)

# s ≠ 1, # s ≠ 0, s ≠ ∅, s.min ∈ s, s.max ∈ s, ¬ s.max < s.min, ¬ (s -- s.min).min ∈ s ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ s ≠ ∅ ∧ s ⊆ s0 → s.min ∈ s0
Proof step 5:
# s ≠ 1, # s ≠ 0, s ≠ ∅, s.max ∈ s, s.min ∈ s, ¬ s.max < s.min, ¬ (s -- s.min).min ∈ s ⊦ s -- s.min ≠ ∅ ∧ s -- s.min ⊆ s

Rule argument

spec: <toplevel>, name:min-subset, seq: ⊦ s ≠ ∅ ∧ s ⊆ s0 → s.min ∈ s0, subst:[s0, s] → [s, s -- s.min], <all paths>

Additional information

Used terms: s, s -- s.min

Proof step 4 from proof bmin-bin (previous step) (back to tree)

⊦ s ≠ ∅ ∧ s ⊆ s0 → s.min ∈ s0

Lemma min-subset

Proof step 5 from proof bmin-bin (previous step) (back to tree)

# s ≠ 1, # s ≠ 0, s ≠ ∅, s.max ∈ s, s.min ∈ s, ¬ s.max < s.min, ¬ (s -- s.min).min ∈ s ⊦ s -- s.min ≠ ∅ ∧ s -- s.min ⊆ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
s.max = s.min, s = ∅ ++ s.min, # s ≠ 1, # s ≠ 0, s ≠ ∅, s.min ∈ s, ¬ (s -- s.min).min ∈ s ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a < a; (s) in-06: ⊦ a ∈ s ∧ s = ∅ ++ b → a = b; (forward) from specification set-basic ⊦ s -- a ⊆ s; (s) ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; (ws)

Proof step 6 from proof bmin-bin (previous step) (back to tree)

s.max = s.min, s = ∅ ++ s.min, # s ≠ 1, # s ≠ 0, s ≠ ∅, s.min ∈ s, ¬ (s -- s.min).min ∈ s ⊦

User applied rule subst equation

And closed the goal!

Rule argument

pos: left 2, flags: discard

Additional information

Used simplifier rules: max-one: ⊦ (∅ ++ a).max = a; min-one: ⊦ (∅ ++ a).min = a; ⊦ #(∅ ++ a) = 1; (s)

Proof Tree for bmin-bin (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.