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Proof step 1 from proof val-trans (conclusion) (back to tree)

⊦ str < str2 ∧ str2 < str3 → str < str3

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof val-trans (previous step) (back to tree)

str < str2, str2 < str3, ¬ str < str3 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 3 from proof val-trans (previous step) (back to tree)

str2 < str3, str < str2, ¬ str < str3 ⊦

User applied rule structural induction

And obtained the following new goals

Rule argument

indvar: str

Additional information

Used simplifier rules: less-empty: ⊦ "" < str ↔ str ≠ ""; Less-notempty: ⊦ ¬ str < ""; string freely generated by "", +;;

Proof step 4 from proof val-trans (previous step) (back to tree)

char + str0 < str2, str2 < str3, ¬ char + str0 < str3, ∀ str2, str3. ¬(str2 < str3 ∧ str0 < str2 ∧ ¬ str0 < str3) ⊦

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 5 from proof val-trans (previous step) (back to tree)

str2 < str3, char ' + str0 < str2, ¬ char ' + str0 < str3, ∀ str, str1. ¬ str < str1 ∨ ¬ str0 < str ∨ str0 < str1 ⊦

User applied rule structural induction

And obtained the following new goals

Rule argument

indvar: str2

Additional information

Used simplifier rules: less-empty: ⊦ "" < str ↔ str ≠ ""; Less-notempty: ⊦ ¬ str < ""; string freely generated by "", +;;

Proof step 6 from proof val-trans (previous step) (back to tree)

char ' + str0 < char0 + str, char0 + str < str3, ¬ char ' + str0 < str3, ∀ str, str1. ¬ str < str1 ∨ ¬ str0 < str ∨ str0 < str1, ∀ str3, char, str0. ¬(str < str3 ∧ char ' + str0 < str ∧ ¬ char ' + str0 < str3 ∧ (∀ str, str1. ¬ str < str1 ∨ ¬ str0 < str ∨ str0 < str1)) ⊦

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 7 from proof val-trans (previous step) (back to tree)

char0 ' + str < str3, char ' + str0 < char0 ' + str, ¬ char ' + str0 < str3, ∀ str, str1. ¬ str < str1 ∨ ¬ str0 < str ∨ str0 < str1, ∀ str1, char1, str2. ¬ str < str1 ∨ ¬ char1 ' + str2 < str ∨ char1 ' + str2 < str1 ∨ ¬(∀ str, str1. ¬ str < str1 ∨ ¬ str2 < str ∨ str2 < str1) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:Less-rec, seq:⊦ char1 ' + str1 < char2 ' + str2 ↔ char1 < char2 ∨ char1 = char2 ∧ str1 < str2, subst:[str2, char2, str1, char1] → [str, char0, str0, char], <all paths>

Additional information

Used terms: str, char0, str0, char

Proof step 8 from proof val-trans (previous step) (back to tree)

⊦ char1 ' + str1 < char2 ' + str2 ↔ char1 < char2 ∨ char1 = char2 ∧ str1 < str2

Lemma Less-rec

Proof step 9 from proof val-trans (previous step) (back to tree)

char0 ' + str < str3, ¬ char ' + str0 < str3, ∀ str1, char1, str2. ¬ str < str1 ∨ ¬ char1 ' + str2 < str ∨ char1 ' + str2 < str1 ∨ ¬(∀ str, str1. ¬ str < str1 ∨ ¬ str2 < str ∨ str2 < str1), ∀ str, str1. ¬ str < str1 ∨ ¬ str0 < str ∨ str0 < str1, char < char0 ∨ char = char0 ∧ str0 < str ⊦

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 5

Additional information

Proof step 10 from proof val-trans (previous step) (back to tree)

char0 ' + str < str3, char < char0, ¬ char ' + str0 < str3, ∀ str, str1. ¬ str < str1 ∨ ¬ str0 < str ∨ str0 < str1, ∀ str1, char1, str2. ¬ str < str1 ∨ ¬ char1 ' + str2 < str ∨ char1 ' + str2 < str1 ∨ ¬(∀ str, str1. ¬ str < str1 ∨ ¬ str2 < str ∨ str2 < str1) ⊦

User applied rule structural induction

And obtained the following new goals

Rule argument

indvar: str3

Additional information

Used simplifier rules: Less-notempty: ⊦ ¬ str < ""; string freely generated by "", +;;

Proof step 11 from proof val-trans (previous step) (back to tree)

char < char0, char0 ' + str < char1 + str1, ¬ char ' + str0 < char1 + str1, ∀ str1, char1, str2. ¬ str < str1 ∨ ¬ char1 ' + str2 < str ∨ char1 ' + str2 < str1 ∨ ¬(∀ str, str1. ¬ str < str1 ∨ ¬ str2 < str ∨ str2 < str1), ∀ str, str1. ¬ str < str1 ∨ ¬ str0 < str ∨ str0 < str1, ∀ char0, str, char, str0. ¬ ( char0 ' + str < str1 ∧ char < char0 ∧ ¬ char ' + str0 < str1 ∧ (∀ str, str1. ¬ str < str1 ∨ ¬ str0 < str ∨ str0 < str1) ∧ (∀ str1, char1, str2. ¬ str < str1 ∨ ¬ char1 ' + str2 < str ∨ char1 ' + str2 < str1 ∨ ¬(∀ str, str1. ¬ str < str1 ∨ ¬ str2 < str ∨ str2 < str1))) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: Less-rec: ⊦ char1 ' + str1 < char2 ' + str2 ↔ char1 < char2 ∨ char1 = char2 ∧ str1 < str2; a: ⊦ (x + y) + z = x + y + z; from specification list[string] ⊦ a + x = a ' + x; (s)

Proof step 12 from proof val-trans (previous step) (back to tree)

char < char0, char0 ' + str < char1 ' + str1, ¬ char < char1, ∀ str, str1. ¬ str < str1 ∨ ¬ str0 < str ∨ str0 < str1, ∀ str1, char1, str2. ¬ str < str1 ∨ ¬ char1 ' + str2 < str ∨ char1 ' + str2 < str1 ∨ ¬(∀ str, str1. ¬ str < str1 ∨ ¬ str2 < str ∨ str2 < str1), ∀ char2, str2, char3, str3. ¬ char2 ' + str2 < str1 ∨ ¬ char3 < char2 ∨ char3 ' + str3 < str1 ∨ ¬(∀ str, str1. ¬ str < str1 ∨ ¬ str3 < str ∨ str3 < str1) ∨ ¬ (∀ str1, char1, str0. ¬ str2 < str1 ∨ ¬ char1 ' + str0 < str2 ∨ char1 ' + str0 < str1 ∨ ¬(∀ str, str1. ¬ str < str1 ∨ ¬ str0 < str ∨ str0 < str1)) ⊦ char = char1 ∧ str0 < str1

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:Less-rec, seq:⊦ char1 ' + str1 < char2 ' + str2 ↔ char1 < char2 ∨ char1 = char2 ∧ str1 < str2, subst:[str2, char2, str1, char1] → [str1, char1, str, char0], <all paths>

Additional information

Used simplifier rules: val-trans-01: ⊦ char < char2 ∧ char2 < char3 → char < char3; Used terms: str1, char1, str, char0

Proof step 13 from proof val-trans (previous step) (back to tree)

⊦ char1 ' + str1 < char2 ' + str2 ↔ char1 < char2 ∨ char1 = char2 ∧ str1 < str2

Lemma Less-rec

Proof step 14 from proof val-trans (previous step) (back to tree)

char = char0, char0 ' + str < str3, str0 < str, ¬ char ' + str0 < str3, ∀ str, str1. ¬ str < str1 ∨ ¬ str0 < str ∨ str0 < str1, ∀ str1, char1, str2. ¬ str < str1 ∨ ¬ char1 ' + str2 < str ∨ char1 ' + str2 < str1 ∨ ¬(∀ str, str1. ¬ str < str1 ∨ ¬ str2 < str ∨ str2 < str1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 15 from proof val-trans (previous step) (back to tree)

str0 < str, char0 ' + str < str3, ¬ char0 ' + str0 < str3, ∀ str1, char1, str2. ¬ str < str1 ∨ ¬ char1 ' + str2 < str ∨ char1 ' + str2 < str1 ∨ ¬(∀ str, str1. ¬ str < str1 ∨ ¬ str2 < str ∨ str2 < str1), ∀ str, str1. ¬ str < str1 ∨ ¬ str0 < str ∨ str0 < str1 ⊦

User applied rule structural induction

And obtained the following new goals

Rule argument

indvar: str3

Additional information

Used simplifier rules: Less-notempty: ⊦ ¬ str < ""; string freely generated by "", +;;

Proof step 16 from proof val-trans (previous step) (back to tree)

char0 ' + str < char + str1, str0 < str, ¬ char0 ' + str0 < char + str1, ∀ str, str1. ¬ str < str1 ∨ ¬ str0 < str ∨ str0 < str1, ∀ str1, char1, str2. ¬ str < str1 ∨ ¬ char1 ' + str2 < str ∨ char1 ' + str2 < str1 ∨ ¬(∀ str, str1. ¬ str < str1 ∨ ¬ str2 < str ∨ str2 < str1), ∀ str0, str, char0. ¬ ( str0 < str ∧ char0 ' + str < str1 ∧ ¬ char0 ' + str0 < str1 ∧ (∀ str1, char1, str2. ¬ str < str1 ∨ ¬ char1 ' + str2 < str ∨ char1 ' + str2 < str1 ∨ ¬(∀ str, str1. ¬ str < str1 ∨ ¬ str2 < str ∨ str2 < str1)) ∧ (∀ str, str1. ¬ str < str1 ∨ ¬ str0 < str ∨ str0 < str1)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: Less-rec: ⊦ char1 ' + str1 < char2 ' + str2 ↔ char1 < char2 ∨ char1 = char2 ∧ str1 < str2; Char-irref: ⊦ ¬ char < char; a: ⊦ (x + y) + z = x + y + z; from specification list[string] ⊦ a + x = a ' + x; (s)

Proof step 17 from proof val-trans (previous step) (back to tree)

str < str1, str0 < str, ¬ str0 < str1, ∀ str2, str3, char1. ¬ str2 < str3 ∨ ¬ char1 ' + str3 < str1 ∨ char1 ' + str2 < str1 ∨ ¬ (∀ str1, char1, str2. ¬ str3 < str1 ∨ ¬ char1 ' + str2 < str3 ∨ char1 ' + str2 < str1 ∨ ¬(∀ str, str1. ¬ str < str1 ∨ ¬ str2 < str ∨ str2 < str1)) ∨ ¬(∀ str, str1. ¬ str < str1 ∨ ¬ str2 < str ∨ str2 < str1), ∀ str, str1. ¬ str < str1 ∨ ¬ str0 < str ∨ str0 < str1, ∀ str1, char1, str2. ¬ str < str1 ∨ ¬ char1 ' + str2 < str ∨ char1 ' + str2 < str1 ∨ ¬(∀ str, str1. ¬ str < str1 ∨ ¬ str2 < str ∨ str2 < str1) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: str, str1

Additional information

Used terms: str, str1

Proof Tree for val-trans (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