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Proof step 1 from proof pis-inf-simp (conclusion) (back to tree)

⊦ πrs(lstep, rstep, lren, rren, ⌜ps '⌝ + ⌜psf⌝ , ⌜paf⌝ ) = lim(⌜λ n. ⌜πrs(lstep, rstep, lren, rren, ps ' + psf.take(n), paf.take(n))⌝ ⌝ )

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof pis-inf-simp (previous step) (back to tree)

πrs(lstep, rstep, lren, rren, ⌜ps '⌝ + ⌜psf⌝ , ⌜paf⌝ ) ≠ lim(⌜λ n. ⌜πrs(lstep, rstep, lren, rren, ps ' + psf.take(n), paf.take(n))⌝ ⌝ ) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream, name: one-inf, seq: ⊦ ⌜a '⌝ + ⌜nf⌝ = ⌜λ n. (n = 0 ⊃ a;nf(n -1))⌝ , subst:[a, nf] → [ps, psf]; ['a] → [('ls, 'rs)], <all paths>

Additional information

Used terms: ps, psf

Proof step 3 from proof pis-inf-simp (previous step) (back to tree)

⊦ ⌜a '⌝ + ⌜nf⌝ = ⌜λ n. (n = 0 ⊃ a;nf(n -1))⌝

A spec-lemma

Proof step 4 from proof pis-inf-simp (previous step) (back to tree)

πrs(lstep, rstep, lren, rren, ⌜λ n. (n = 0 ⊃ ps;psf(n -1))⌝ , ⌜paf⌝ ) ≠ lim(⌜λ n. ⌜πrs(lstep, rstep, lren, rren, ps ' + psf.take(n), paf.take(n))⌝ ⌝ ) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ n -1 = n - 1; (s)

Proof step 5 from proof pis-inf-simp (previous step) (back to tree)

πrs(lstep, rstep, lren, rren, ⌜λ n. (n = 0 ⊃ ps;psf(n - 1))⌝ , ⌜paf⌝ ) ≠ lim(⌜λ n. ⌜πrs(lstep, rstep, lren, rren, ps ' + psf.take(n), paf.take(n))⌝ ⌝ ) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name: projstate-inf-def, seq:⊦ πrs(lstep, rstep, lren, rren, ⌜psf⌝ , ⌜paf⌝ ) = lim(⌜λ n. ⌜πrs(lstep, rstep, lren, rren, psf.take(n + 1), paf.take(n))⌝ ⌝ ), subst:[psf, lstep, paf, rren, lren, rstep] → [λ n. (n = 0 ⊃ ps;psf(n - 1)), lstep, paf, rren, lren, rstep]; ['ls, 'ra, 'la, 'pa, 'rs] → ['ls, 'ra, 'la, 'pa, 'rs], <all paths>

Additional information

Used terms: paf, rstep, lren, rren, lstep, λ n. (n = 0 ⊃ ps;psf(n - 1))

Proof step 6 from proof pis-inf-simp (previous step) (back to tree)

⊦ πrs(lstep, rstep, lren, rren, ⌜psf⌝ , ⌜paf⌝ ) = lim(⌜λ n. ⌜πrs(lstep, rstep, lren, rren, psf.take(n + 1), paf.take(n))⌝ ⌝ )

Lemma projstate-inf-def

Proof step 7 from proof pis-inf-simp (previous step) (back to tree)

lim(⌜λ n. ⌜πrs(lstep, rstep, lren, rren, (λ n. (n = 0 ⊃ ps;psf(n - 1))).take(n + 1), paf.take(n))⌝ ⌝ ) ≠ lim(⌜λ n. ⌜πrs(lstep, rstep, lren, rren, ps ' + psf.take(n), paf.take(n))⌝ ⌝ ) ⊦

User applied rule reduce equation

And obtained the following new goals

Rule argument

pos: left 1

Additional information

Proof step 8 from proof pis-inf-simp (previous step) (back to tree)

⌜λ n. ⌜πrs(lstep, rstep, lren, rren, (λ n. (n = 0 ⊃ ps;psf(n - 1))).take(n + 1), paf.take(n))⌝ ⌝ ≠ ⌜λ n. ⌜πrs(lstep, rstep, lren, rren, ps ' + psf.take(n), paf.take(n))⌝ ⌝ ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ = ⌜x0⌝ ↔ x = x0; (s) ⊦ ⌜nf⌝ = ⌜nf0⌝ ↔ nf = nf0; (s)

Proof step 9 from proof pis-inf-simp (previous step) (back to tree)

πrs(lstep, rstep, lren, rren, (λ n. (n = 0 ⊃ ps;psf(n - 1))).take(n + 1), paf.take(n)) ≠ πrs(lstep, rstep, lren, rren, ps ' + psf.take(n), paf.take(n)) ⊦

User applied rule reduce equation

And obtained the following new goals

Rule argument

pos: left 1

Additional information

Proof step 10 from proof pis-inf-simp (previous step) (back to tree)

(λ n. (n = 0 ⊃ ps;psf(n - 1))).take(n + 1) ≠ ps ' + psf.take(n) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: stream, name: to-succ-nf, seq: ⊦ nf.take(n + 1) = nf(0) + (λ n. nf(n + 1)).take(n), subst:[nf, n] → [λ n. (n = 0 ⊃ ps;psf(n - 1)), n]; ['a] → [('ls, 'rs)], <all paths>

Additional information

Used simplifier rules: ⊦ m + 1 ≠ 0; (s) ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s) ⊦ n + 0 = n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ a + x = a ' + x; (s) ⊦ # x = # z → (x + y = z + y0 ↔ x = z ∧ y = y0); (ws) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: λ n. (n = 0 ⊃ ps;psf(n - 1)), n

Proof step 11 from proof pis-inf-simp (previous step) (back to tree)

⊦ nf.take(n + 1) = nf(0) + (λ n. nf(n + 1)).take(n)

A spec-lemma

Proof Tree for pis-inf-simp (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