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Proof step 1 from proof plst-01 (conclusion) (back to tree)

r + r1 < r0 + r2, ¬ r4 < r2, ¬ r1 < r3 ⊦ r + r3 < r0 + r4

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof plst-01 (previous step) (back to tree)

r + r1 < r0 + r2, ¬ r + r3 < r0 + r4, ¬ r4 < r2, ¬ r1 < r3 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: l: ⊦ r < r0 ↔ ¬(r = r0 ∨ r0 < r);

Proof step 3 from proof plst-01 (previous step) (back to tree)

r + r1 < r0 + r2 ⊦ r + r3 ≠ r0 + r4 ∧ ¬ r0 + r4 < r + r3, r4 ≠ r2 ∧ ¬ r2 < r4, r1 ≠ r3 ∧ ¬ r3 < r1

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 3

Additional information

Proof step 4 from proof plst-01 (previous step) (back to tree)

r1 = r3, r + r1 < r0 + r2 ⊦ r4 ≠ r2 ∧ ¬ r2 < r4, r + r3 ≠ r0 + r4 ∧ ¬ r0 + r4 < r + r3

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 5 from proof plst-01 (previous step) (back to tree)

r + r3 < r0 + r2 ⊦ r + r3 ≠ r0 + r4 ∧ ¬ r0 + r4 < r + r3, r4 ≠ r2 ∧ ¬ r2 < r4

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: bls-01: ⊦ r < r0 → ¬ r0 < r; less-add-cancel-l: ⊦ r + r0 < r + r1 ↔ r0 < r1; irref: ⊦ ¬ r < r; add-assoc: ⊦ (r + r0) + r1 = r + r0 + r1; add-comm: ⊦ r + r0 = r0 + r; trans: ⊦ r < r0 ∧ r0 < r1 → r < r1; l: ⊦ r < r0 ↔ ¬(r = r0 ∨ r0 < r); less-add-cancel-l: ⊦ r + r0 < r + r1 ↔ r0 < r1; bls-01: ⊦ r < r0 → ¬ r0 < r; add-assoc: ⊦ (r + r0) + r1 = r + r0 + r1; add-comm: ⊦ r + r0 = r0 + r;

Proof step 6 from proof plst-01 (previous step) (back to tree)

r3 < r1, r + r1 < r0 + r2 ⊦ r4 ≠ r2 ∧ ¬ r2 < r4, r + r3 ≠ r0 + r4 ∧ ¬ r0 + r4 < r + r3

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Proof step 7 from proof plst-01 (previous step) (back to tree)

r4 = r2, r + r1 < r0 + r2, r3 < r1 ⊦ r + r3 ≠ r0 + r4 ∧ ¬ r0 + r4 < r + r3

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: trans: ⊦ r < r0 ∧ r0 < r1 → r < r1; less-add-cancel-l: ⊦ r + r0 < r + r1 ↔ r0 < r1; bls-01: ⊦ r < r0 → ¬ r0 < r; add-assoc: ⊦ (r + r0) + r1 = r + r0 + r1; add-comm: ⊦ r + r0 = r0 + r; l: ⊦ r < r0 ↔ ¬(r = r0 ∨ r0 < r);

Proof step 8 from proof plst-01 (previous step) (back to tree)

r + r3 = r0 + r2, r + r1 < r0 + r2, r3 < r1 ⊦

User applied rule subst equation

And closed the goal!

Rule argument

pos: left 1, flags: rotate, discard

Additional information

Used simplifier rules: less-add-cancel-l: ⊦ r + r0 < r + r1 ↔ r0 < r1; bls-01: ⊦ r < r0 → ¬ r0 < r; add-assoc: ⊦ (r + r0) + r1 = r + r0 + r1; add-comm: ⊦ r + r0 = r0 + r;

Proof step 9 from proof plst-01 (previous step) (back to tree)

r2 < r4, r + r1 < r0 + r2, r3 < r1 ⊦ r + r3 ≠ r0 + r4 ∧ ¬ r0 + r4 < r + r3

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: r + r3 < r0 + r2

Additional information

Used simplifier rules: lst-04: ⊦ r + r0 < r1 ∧ r2 < r0 → r + r2 < r1; add-assoc: ⊦ (r + r0) + r1 = r + r0 + r1; add-comm: ⊦ r + r0 = r0 + r; Used formulas: r + r3 < r0 + r2

Proof step 10 from proof plst-01 (previous step) (back to tree)

r3 < r1, r + r1 < r0 + r2, r2 < r4, r + r3 < r0 + r2 ⊦ r + r3 ≠ r0 + r4 ∧ ¬ r0 + r4 < r + r3

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: r + r3 < r0 + r4

Additional information

Used formulas: r + r3 < r0 + r4

Proof step 11 from proof plst-01 (previous step) (back to tree)

r + r3 < r0 + r2, r2 < r4, r + r1 < r0 + r2, r3 < r1, r + r3 < r0 + r4 ⊦ r + r3 ≠ r0 + r4 ∧ ¬ r0 + r4 < r + r3

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: lst-04: ⊦ r + r0 < r1 ∧ r2 < r0 → r + r2 < r1; less-add-cancel-l: ⊦ r + r0 < r + r1 ↔ r0 < r1; irref: ⊦ ¬ r < r; add-assoc: ⊦ (r + r0) + r1 = r + r0 + r1; add-comm: ⊦ r + r0 = r0 + r; lst-04: ⊦ r + r0 < r1 ∧ r2 < r0 → r + r2 < r1; add-assoc: ⊦ (r + r0) + r1 = r + r0 + r1; add-comm: ⊦ r + r0 = r0 + r; bls-01: ⊦ r < r0 → ¬ r0 < r;

Proof step 12 from proof plst-01 (previous step) (back to tree)

r + r3 < r0 + r2, r2 < r4, r + r1 < r0 + r2, r3 < r1, ¬ r + r3 < r0 + r4 ⊦ r + r3 ≠ r0 + r4 ∧ ¬ r0 + r4 < r + r3

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name: trans, seq: ⊦ r < r0 ∧ r0 < r1 → r < r1, subst:[r0, r, r1] → [r0 + r2, r + r3, r0 + r4], <all paths>

Additional information

Used simplifier rules: l: ⊦ r < r0 ↔ ¬(r = r0 ∨ r0 < r); lst-04: ⊦ r + r0 < r1 ∧ r2 < r0 → r + r2 < r1; less-add-cancel-l: ⊦ r + r0 < r + r1 ↔ r0 < r1; add-assoc: ⊦ (r + r0) + r1 = r + r0 + r1; add-comm: ⊦ r + r0 = r0 + r; Used terms: r0 + r2, r + r3, r0 + r4

Proof step 13 from proof plst-01 (previous step) (back to tree)

⊦ r < r0 ∧ r0 < r1 → r < r1

Lemma trans

Proof Tree for plst-01 (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