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Proof step 1 from proof inffrag-rec1 (conclusion) (back to tree)

⊦ fragω(IOAut(start, step, ext))(s ' + sf, af) ↔ step(s, af(0), sf(0)) ∧ fragω(IOAut(start, step, ext))(sf, λ n. af(n + 1))

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof inffrag-rec1 (previous step) (back to tree)

¬(fragω(IOAut(start, step, ext))(s ' + sf, af) ↔ step(s, af(0), sf(0)) ∧ fragω(IOAut(start, step, ext))(sf, λ n. af(n + 1))) ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:inffrag-rec, seq:⊦ fragω(IOAut(start, step, ext))(s ' + sf, a ' + af) ↔ step(s, a, sf(0)) ∧ fragω(IOAut(start, step, ext))(sf, af), subst:[a, step, s, start, sf, ext, af] → [af(0), step, s, start, sf, ext, λ n. af(n + 1)]; ['s, 'a, 'e] → ['s, 'a, 'e]

Additional information

Used terms: sf, ext, s, af(0), step, λ n. af(n + 1), start

Proof step 3 from proof inffrag-rec1 (previous step) (back to tree)

⊦ fragω(IOAut(start, step, ext))(s ' + sf, a ' + af) ↔ step(s, a, sf(0)) ∧ fragω(IOAut(start, step, ext))(sf, af)

Lemma inffrag-rec

Proof step 4 from proof inffrag-rec1 (previous step) (back to tree)

fragω(IOAut(start, step, ext))(s ' + sf, af(0) ' + (λ n. af(n + 1))) ↔ step(s, af(0), sf(0)) ∧ fragω(IOAut(start, step, ext))(sf, λ n. af(n + 1)), ¬(fragω(IOAut(start, step, ext))(s ' + sf, af) ↔ step(s, af(0), sf(0)) ∧ fragω(IOAut(start, step, ext))(sf, λ n. af(n + 1))) ⊦

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: af(0) ' + (λ n. af(n + 1)) = af

Additional information

Used formulas: af(0) ' + (λ n. af(n + 1)) = af

Proof step 5 from proof inffrag-rec1 (previous step) (back to tree)

fragω(IOAut(start, step, ext))(s ' + sf, af(0) ' + (λ n. af(n + 1))) ↔ step(s, af(0), sf(0)) ∧ fragω(IOAut(start, step, ext))(sf, λ n. af(n + 1)), af(0) ' + (λ n. af(n + 1)) ≠ af, ¬(fragω(IOAut(start, step, ext))(s ' + sf, af) ↔ step(s, af(0), sf(0)) ∧ fragω(IOAut(start, step, ext))(sf, λ n. af(n + 1))) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream, name:add-list-fun-def, seq: ⊦ x + nf = (λ n. (n < # x ⊃ x[n];nf(n - # x))), subst:[x, nf] → [af(0) ', λ n. af(n + 1)]; ['a] → ['a], <all paths>

Additional information

Used terms: af(0) ', λ n. af(n + 1)

Proof step 6 from proof inffrag-rec1 (previous step) (back to tree)

⊦ x + nf = (λ n. (n < # x ⊃ x[n];nf(n - # x)))

A spec-lemma

Proof step 7 from proof inffrag-rec1 (previous step) (back to tree)

fragω(IOAut(start, step, ext))(s ' + sf, λ n. (n < # af(0) ' ⊃ af(0) '[n];af((n - # af(0) ') + 1))) ↔ step(s, af(0), sf(0)) ∧ fragω(IOAut(start, step, ext))(sf, λ n. af(n + 1)), ¬(fragω(IOAut(start, step, ext))(s ' + sf, af) ↔ step(s, af(0), sf(0)) ∧ fragω(IOAut(start, step, ext))(sf, λ n. af(n + 1))), (λ n. (n < # af(0) ' ⊃ af(0) '[n];af((n - # af(0) ') + 1))) ≠ af ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ # a ' = 0 +1; (s) ⊦ a '[0] = a; (s) ⊦ m < 1 ↔ m = 0; (s)

Proof step 8 from proof inffrag-rec1 (previous step) (back to tree)

fragω(IOAut(start, step, ext))(s ' + sf, λ n. (n = 0 ⊃ af(0);af((n - 1) + 1))) ↔ step(s, af(0), sf(0)) ∧ fragω(IOAut(start, step, ext))(sf, λ n. af(n + 1)), (n = 0 ⊃ af(0);af((n - 1) + 1)) ≠ af(n), ¬(fragω(IOAut(start, step, ext))(s ' + sf, af) ↔ step(s, af(0), sf(0)) ∧ fragω(IOAut(start, step, ext))(sf, λ n. af(n + 1))) ⊦

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 2

Additional information

Proof step 9 from proof inffrag-rec1 (previous step) (back to tree)

fragω(IOAut(start, step, ext))(s ' + sf, λ n. (n = 0 ⊃ af(0);af((n - 1) + 1))) ↔ step(s, af(0), sf(0)) ∧ fragω(IOAut(start, step, ext))(sf, λ n. af(n + 1)), ¬(fragω(IOAut(start, step, ext))(s ' + sf, af) ↔ step(s, af(0), sf(0)) ∧ fragω(IOAut(start, step, ext))(sf, λ n. af(n + 1))), af((n - 1) + 1) ≠ af(n), n ≠ 0 ⊦

Heuristic elimination applied rule apply elim lemma

And closed the goal!

Rule argument

spec: nat, name:elim-pred-c, seq:n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1, subst:[n, m] → [n, m]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m

Proof Tree for inffrag-rec1 (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