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

⊦ a ∈ ax ↔ (∃ d, ax0, ax1. ax = ax0 + (a, d) ' + ax1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

⊦ a ∈ ax ↔ (∃ d, ax0, ax1. ax = ax0 + (a, d) ' + ax1)

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:inassoc, seq: ⊦ a ∈ ax ↔ (∃ d. (a, d) ∈ ax), subst:[ax, a] → [ax, a], <all paths>

Additional information

Used terms: ax, a

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

⊦ a ∈ ax ↔ (∃ d. (a, d) ∈ ax)

Lemma inassoc

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

⊦ (∃ d. (a, d) ∈ ax) ↔ (∃ d, ax0, ax1. ax = ax0 + (a, d) ' + ax1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

⊦ (∃ d. (a, d) ∈ ax) ↔ (∃ d, ax0, ax1. ax = ax0 + (a, d) ' + ax1)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

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

∃ d. (a, d) ∈ ax ⊦ ∃ d, ax0, ax1. ax = ax0 + (a, d) ' + ax1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

(a, d) ∈ ax ⊦ ∃ d, ax0, ax1. ax = ax0 + (a, d) ' + ax1

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: polylist, name:in-def, seq: ⊦ a ∈ x ↔ (∃ y, z. x = y + (a + z)), subst:[x, a] → [ax, (a, d)]; ['a] → [('a, 'b)], <all paths>

Additional information

Used terms: ax, (a, d)

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

⊦ a ∈ x ↔ (∃ y, z. x = y + (a + z))

A spec-lemma

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

∃ y, z. ax = y + ((a, d) + z) ⊦ ∃ d, ax0, ax1. ax = ax0 + (a, d) ' + ax1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s)

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

⊦ ∃ d0, ax2, ax3. ax0 + (a, d) ' + ax1 = ax2 + (a, d0) ' + ax3

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: d, ax0, ax1

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification polylist Used terms: d, ax0, ax1

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

∃ d, ax0, ax1. ax = ax0 + (a, d) ' + ax1 ⊦ ∃ d. (a, d) ∈ ax

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification polylist ⊦ a ∈ a0 ' ↔ a = a0; (s) ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)

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

⊦ ∃ d0. (a, d0) ∈ ax0 ∨ d0 = d ∨ (a, d0) ∈ ax1

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: d

Additional information

Used terms: d

Proof Tree for ex (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