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Proof step 1 from proof Sep-eq (conclusion) (back to tree)

⊦ (∀ a. a ∈ st → adP(a, st(a)) = adP0(a, st(a))) → adP ⊗_ st = adP0 ⊗_ st

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof Sep-eq (previous step) (back to tree)

adP ⊗_ st ≠ adP0 ⊗_ st, ∀ a. a ∈ st → adP(a, st(a)) = adP0(a, st(a)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 3 from proof Sep-eq (previous step) (back to tree)

adP ⊗_ st ≠ adP0 ⊗_ st, ∀ a. a ∈ st → adP(a, st(a)) = adP0(a, st(a)) ⊦

User applied rule expand left

And obtained the following new goals

Rule argument

pos: left 1

Additional information

Proof step 4 from proof Sep-eq (previous step) (back to tree)

¬((adP ⊗_ st)(h) ↔ (adP0 ⊗_ st)(h)), ∀ a. a ∈ st → adP(a, st(a)) = adP0(a, st(a)) ⊦

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information

Proof step 5 from proof Sep-eq (previous step) (back to tree)

(adP ⊗_ st)(h), ¬ (adP0 ⊗_ st)(h), ∀ a. a ∈ st → adP(a, st(a)) = adP0(a, st(a)) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name: Sep-forall, seq:∀ a, h. a ∈ st ∧ adP0(a, st(a))(h) → adP(a, st(a))(h) ⊦ (adP0 ⊗_ st)(h) → (adP ⊗_ st)(h), subst:[adP0, h, st, adP] → [adP, h, st, adP0], <all paths>

Additional information

Used terms: adP, h, st, adP0

Proof step 6 from proof Sep-eq (previous step) (back to tree)

∀ a, h. a ∈ st ∧ adP0(a, st(a))(h) → adP(a, st(a))(h) ⊦ (adP0 ⊗_ st)(h) → (adP ⊗_ st)(h)

Lemma Sep-forall

Proof step 7 from proof Sep-eq (previous step) (back to tree)

(adP ⊗_ st)(h), ¬ (adP0 ⊗_ st)(h), ∀ a. a ∈ st → adP(a, st(a)) = adP0(a, st(a)) ⊦ (∀ a, h. a ∈ st ∧ adP(a, st(a))(h) → adP0(a, st(a))(h)) ∧ (adP ⊗_ st)(h)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 8 from proof Sep-eq (previous step) (back to tree)

a ∈ st, (adP ⊗_ st)(h), adP(a, st(a))(h0), ¬ (adP0 ⊗_ st)(h), ¬ adP0(a, st(a))(h0), ∀ a. a ∈ st → adP(a, st(a)) = adP0(a, st(a)) ⊦

User applied rule all left

And closed the goal!

Rule argument

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

Additional information

Used terms: a

Proof step 9 from proof Sep-eq (previous step) (back to tree)

(adP0 ⊗_ st)(h), ¬ (adP ⊗_ st)(h), ∀ a. a ∈ st → adP(a, st(a)) = adP0(a, st(a)) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name: Sep-forall, seq:∀ a, h. a ∈ st ∧ adP0(a, st(a))(h) → adP(a, st(a))(h) ⊦ (adP0 ⊗_ st)(h) → (adP ⊗_ st)(h), subst:[adP0, h, st, adP] → [adP0, h, st, adP], <all paths>

Additional information

Used terms: adP0, h, st, adP

Proof step 10 from proof Sep-eq (previous step) (back to tree)

∀ a, h. a ∈ st ∧ adP0(a, st(a))(h) → adP(a, st(a))(h) ⊦ (adP0 ⊗_ st)(h) → (adP ⊗_ st)(h)

Lemma Sep-forall

Proof step 11 from proof Sep-eq (previous step) (back to tree)

(adP0 ⊗_ st)(h), ¬ (adP ⊗_ st)(h), ∀ a. a ∈ st → adP(a, st(a)) = adP0(a, st(a)) ⊦ (∀ a, h. a ∈ st ∧ adP0(a, st(a))(h) → adP(a, st(a))(h)) ∧ (adP0 ⊗_ st)(h)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 12 from proof Sep-eq (previous step) (back to tree)

a ∈ st, (adP0 ⊗_ st)(h), adP0(a, st(a))(h0), ¬ (adP ⊗_ st)(h), ¬ adP(a, st(a))(h0), ∀ a. a ∈ st → adP(a, st(a)) = adP0(a, st(a)) ⊦

User applied rule all left

And closed the goal!

Rule argument

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

Additional information

Used terms: a

Proof Tree for Sep-eq (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