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

⊦ bag0 ⊆ bag1 ∧ a ∈ bag0 → (a ∈ bag1 ↔ true)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

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

¬(a ∈ bag1 ↔ true), a ∈ bag0, bag0 ⊆ bag1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

a ∈ bag0, bag0 ⊆ bag1, ¬ a ∈ bag1 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:count-subset, seq: ⊦ bag0 ⊆ bag1 ↔ (∀ a. count(a, bag0) ≤ count(a, bag1)), subst:[bag1, bag0] → [bag1, bag0], <all paths>

Additional information

Used terms: bag1, bag0

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

⊦ bag0 ⊆ bag1 ↔ (∀ a. count(a, bag0) ≤ count(a, bag1))

Lemma count-subset

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

a ∈ bag0, ¬ a ∈ bag1, ∀ a. count(a, bag0) ≤ count(a, bag1) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:in, seq: ⊦ a ∈ bag ↔ count(a, bag) ≠ 0, subst:[bag, a] → [bag0, a], <all paths>

Additional information

Used terms: bag0, a

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

⊦ a ∈ bag ↔ count(a, bag) ≠ 0

Lemma in

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

count(a, bag0) ≠ 0, ¬ a ∈ bag1, ∀ a. count(a, bag0) ≤ count(a, bag1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

count(a, bag0) ≠ 0, ¬ a ∈ bag1, ∀ a. count(a, bag0) ≤ count(a, bag1) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:in, seq: ⊦ a ∈ bag ↔ count(a, bag) ≠ 0, subst:[bag, a] → [bag1, a], <all paths>

Additional information

Used terms: bag1, a

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

⊦ a ∈ bag ↔ count(a, bag) ≠ 0

Lemma in

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

count(a, bag1) = 0, count(a, bag0) ≠ 0, ∀ a. count(a, bag0) ≤ count(a, bag1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

count(a, bag1) = 0, count(a, bag0) ≠ 0, ∀ a. count(a, bag0) ≤ count(a, bag1) ⊦

User applied rule all left

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: ⊦ m ≤ 0 ↔ m = 0; (s) Used terms: a

Proof Tree for subset-elem (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