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Proof step 1 from proof wflex (conclusion) (back to tree)

∀ a, b. (∀ a0, b0. lex<(a0, b0, a, b) → iHPQ(a0, b0)) → iHPQ(a, b) ⊦ iHPQ(a, b)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof wflex (previous step) (back to tree)

¬ iHPQ(a, b), ∀ a, b. (∀ a0, b0. lex<(a0, b0, a, b) → iHPQ(a0, b0)) → iHPQ(a, b) ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: wellfounded[lexical-param], name:wellfounded, seq:∀ a. (∀ a0. a0 ≪1 a → iHP(a0)) → iHP(a) ⊦ iHP(a), subst:[iHP, a] → [λ a. ∀ b. iHPQ(a, b), a]

Additional information

Used terms: λ a. ∀ b. iHPQ(a, b), a

Proof step 3 from proof wflex (previous step) (back to tree)

∀ a. (∀ a0. a0 ≪1 a → iHP(a0)) → iHP(a) ⊦ iHP(a)

A spec-lemma

Proof step 4 from proof wflex (previous step) (back to tree)

¬ iHPQ(a, b), ∀ a, b. (∀ a0, b0. lex<(a0, b0, a, b) → iHPQ(a0, b0)) → iHPQ(a, b) ⊦ ∀ a. (∀ a0. a0 ≪1 a → (∀ b. iHPQ(a0, b))) → (∀ b. iHPQ(a, b))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: lex-def: ⊦ lex<(a0, b0, a, b) ↔ a0 ≪1 a ∨ a0 = a ∧ b0 ≪2 b;

Proof step 5 from proof wflex (previous step) (back to tree)

¬ iHPQ(a0, b0), ¬ iHPQ(a, b), ∀ a1. a1 ≪1 a0 → (∀ b. iHPQ(a1, b)), ∀ a1, b1. (∀ a2, b2. a2 ≪1 a1 ∨ a2 = a1 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a1, b1) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 4 terms: a0, b0 abort: F discard: F computed: T

Additional information

Used terms: a0, b0

Proof step 6 from proof wflex (previous step) (back to tree)

¬ iHPQ(a0, b0), ¬ iHPQ(a, b), (∀ a2, b2. a2 ≪1 a0 ∨ a2 = a0 ∧ b2 ≪2 b0 → iHPQ(a2, b2)) → iHPQ(a0, b0), ∀ a1, b1. (∀ a2, b2. a2 ≪1 a1 ∨ a2 = a1 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a1, b1), ∀ a1. a1 ≪1 a0 → (∀ b. iHPQ(a1, b)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 7 from proof wflex (previous step) (back to tree)

¬ iHPQ(a0, b0), ¬ iHPQ(a, b), ¬ iHPQ(a2, b2), ∀ a1. a1 ≪1 a0 → (∀ b. iHPQ(a1, b)), ∀ a1, b1. (∀ a2, b2. a2 ≪1 a1 ∨ a2 = a1 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a1, b1), a2 ≪1 a0 ∨ a2 = a0 ∧ b2 ≪2 b0 ⊦

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 6

Additional information

Proof step 8 from proof wflex (previous step) (back to tree)

a2 ≪1 a0, ¬ iHPQ(a0, b0), ¬ iHPQ(a, b), ¬ iHPQ(a2, b2), ∀ a1, b1. (∀ a2, b2. a2 ≪1 a1 ∨ a2 = a1 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a1, b1), ∀ a1. a1 ≪1 a0 → (∀ b. iHPQ(a1, b)) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 6 terms: a2 abort: F discard: T computed: T

Additional information

Used terms: a2

Proof step 9 from proof wflex (previous step) (back to tree)

a2 ≪1 a0, ¬ iHPQ(a0, b0), ¬ iHPQ(a, b), ¬ iHPQ(a2, b2), a2 ≪1 a0 → (∀ b. iHPQ(a2, b)), ∀ a1, b1. (∀ a2, b2. a2 ≪1 a1 ∨ a2 = a1 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a1, b1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 10 from proof wflex (previous step) (back to tree)

a2 ≪1 a0, ¬ iHPQ(a0, b0), ¬ iHPQ(a, b), ¬ iHPQ(a2, b2), ∀ a1, b1. (∀ a2, b2. a2 ≪1 a1 ∨ a2 = a1 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a1, b1), ∀ b. iHPQ(a2, b) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: b2

Additional information

Used terms: b2

Proof step 11 from proof wflex (previous step) (back to tree)

a2 = a0, b2 ≪2 b0, ¬ iHPQ(a0, b0), ¬ iHPQ(a, b), ¬ iHPQ(a2, b2), ∀ a1, b1. (∀ a2, b2. a2 ≪1 a1 ∨ a2 = a1 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a1, b1), ∀ a1. a1 ≪1 a0 → (∀ b. iHPQ(a1, b)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 12 from proof wflex (previous step) (back to tree)

b2 ≪2 b0, ¬ iHPQ(a0, b0), ¬ iHPQ(a, b), ¬ iHPQ(a0, b2), ∀ a1. a1 ≪1 a0 → (∀ b. iHPQ(a1, b)), ∀ a1, b1. (∀ a2, b2. a2 ≪1 a1 ∨ a2 = a1 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a1, b1) ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: wellfounded[lexical-param1], name:wellfounded, seq:∀ a. (∀ a0. a0 ≪2 a → iHP(a0)) → iHP(a) ⊦ iHP(a), subst:[iHP, a] → [λ b. iHPQ(a0, b), b0]

Additional information

Used terms: λ b. iHPQ(a0, b), b0

Proof step 13 from proof wflex (previous step) (back to tree)

∀ a. (∀ a0. a0 ≪2 a → iHP(a0)) → iHP(a) ⊦ iHP(a)

A spec-lemma

Proof step 14 from proof wflex (previous step) (back to tree)

b2 ≪2 b0, ¬ iHPQ(a0, b0), ¬ iHPQ(a, b), ¬ iHPQ(a0, b2), ∀ a1, b1. (∀ a2, b2. a2 ≪1 a1 ∨ a2 = a1 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a1, b1), ∀ a1. a1 ≪1 a0 → (∀ b. iHPQ(a1, b)) ⊦ ∀ b. (∀ b0. b0 ≪2 b → iHPQ(a0, b0)) → iHPQ(a0, b)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 15 from proof wflex (previous step) (back to tree)

b2 ≪2 b0, ¬ iHPQ(a0, b1), ¬ iHPQ(a0, b0), ¬ iHPQ(a, b), ¬ iHPQ(a0, b2), ∀ b0. b0 ≪2 b1 → iHPQ(a0, b0), ∀ a1. a1 ≪1 a0 → (∀ b. iHPQ(a1, b)), ∀ a1, b1. (∀ a2, b2. a2 ≪1 a1 ∨ a2 = a1 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a1, b1) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 8 terms: a0, b1 abort: F discard: F computed: T

Additional information

Used terms: a0, b1

Proof step 16 from proof wflex (previous step) (back to tree)

b2 ≪2 b0, ¬ iHPQ(a0, b1), ¬ iHPQ(a0, b0), ¬ iHPQ(a, b), ¬ iHPQ(a0, b2), (∀ a2, b2. a2 ≪1 a0 ∨ a2 = a0 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a0, b1), ∀ a1, b1. (∀ a2, b2. a2 ≪1 a1 ∨ a2 = a1 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a1, b1), ∀ a1. a1 ≪1 a0 → (∀ b. iHPQ(a1, b)), ∀ b0. b0 ≪2 b1 → iHPQ(a0, b0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 17 from proof wflex (previous step) (back to tree)

b2 ≪2 b0, ¬ iHPQ(a0, b1), ¬ iHPQ(a0, b0), ¬ iHPQ(a, b), ¬ iHPQ(a0, b2), ¬ iHPQ(a2, b3), ∀ b0. b0 ≪2 b1 → iHPQ(a0, b0), ∀ a1. a1 ≪1 a0 → (∀ b. iHPQ(a1, b)), ∀ a1, b1. (∀ a2, b2. a2 ≪1 a1 ∨ a2 = a1 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a1, b1), a2 ≪1 a0 ∨ a2 = a0 ∧ b3 ≪2 b1 ⊦

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 10

Additional information

Proof step 18 from proof wflex (previous step) (back to tree)

a2 ≪1 a0, b2 ≪2 b0, ¬ iHPQ(a0, b1), ¬ iHPQ(a0, b0), ¬ iHPQ(a, b), ¬ iHPQ(a0, b2), ¬ iHPQ(a2, b3), ∀ a1, b1. (∀ a2, b2. a2 ≪1 a1 ∨ a2 = a1 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a1, b1), ∀ a1. a1 ≪1 a0 → (∀ b. iHPQ(a1, b)), ∀ b0. b0 ≪2 b1 → iHPQ(a0, b0) ⊦

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 9 terms: a2 abort: F discard: T computed: T

Additional information

Used terms: a2

Proof step 19 from proof wflex (previous step) (back to tree)

a2 ≪1 a0, b2 ≪2 b0, ¬ iHPQ(a0, b1), ¬ iHPQ(a0, b0), ¬ iHPQ(a, b), ¬ iHPQ(a0, b2), ¬ iHPQ(a2, b3), a2 ≪1 a0 → (∀ b. iHPQ(a2, b)), ∀ b0. b0 ≪2 b1 → iHPQ(a0, b0), ∀ a1, b1. (∀ a2, b2. a2 ≪1 a1 ∨ a2 = a1 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a1, b1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 20 from proof wflex (previous step) (back to tree)

a2 ≪1 a0, b2 ≪2 b0, ¬ iHPQ(a0, b1), ¬ iHPQ(a0, b0), ¬ iHPQ(a, b), ¬ iHPQ(a0, b2), ¬ iHPQ(a2, b3), ∀ a1, b1. (∀ a2, b2. a2 ≪1 a1 ∨ a2 = a1 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a1, b1), ∀ b0. b0 ≪2 b1 → iHPQ(a0, b0), ∀ b. iHPQ(a2, b) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 10 terms: b3

Additional information

Used terms: b3

Proof step 21 from proof wflex (previous step) (back to tree)

a2 = a0, b2 ≪2 b0, b3 ≪2 b1, ¬ iHPQ(a0, b1), ¬ iHPQ(a0, b0), ¬ iHPQ(a, b), ¬ iHPQ(a0, b2), ¬ iHPQ(a2, b3), ∀ a1, b1. (∀ a2, b2. a2 ≪1 a1 ∨ a2 = a1 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a1, b1), ∀ a1. a1 ≪1 a0 → (∀ b. iHPQ(a1, b)), ∀ b0. b0 ≪2 b1 → iHPQ(a0, b0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 22 from proof wflex (previous step) (back to tree)

b3 ≪2 b1, b2 ≪2 b0, ¬ iHPQ(a0, b1), ¬ iHPQ(a0, b0), ¬ iHPQ(a, b), ¬ iHPQ(a0, b2), ¬ iHPQ(a0, b3), ∀ b0. b0 ≪2 b1 → iHPQ(a0, b0), ∀ a1. a1 ≪1 a0 → (∀ b. iHPQ(a1, b)), ∀ a1, b1. (∀ a2, b2. a2 ≪1 a1 ∨ a2 = a1 ∧ b2 ≪2 b1 → iHPQ(a2, b2)) → iHPQ(a1, b1) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 8 terms: b3

Additional information

Used terms: b3

Proof step 23 from proof wflex (previous step) (back to tree)

¬ iHPQ(a, b), ∀ a, b. (∀ a0, b0. lex<(a0, b0, a, b) → iHPQ(a0, b0)) → iHPQ(a, b), ∀ b. iHPQ(a, b) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: lex-def: ⊦ lex<(a0, b0, a, b) ↔ a0 ≪1 a ∨ a0 = a ∧ b0 ≪2 b;

Proof step 24 from proof wflex (previous step) (back to tree)

¬ iHPQ(a, b), ∀ b. iHPQ(a, b), ∀ a0, b0. (∀ a1, b1. a1 ≪1 a0 ∨ a1 = a0 ∧ b1 ≪2 b0 → iHPQ(a1, b1)) → iHPQ(a0, b0) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: b

Additional information

Used terms: b

Proof Tree for wflex (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