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Proof step 2 from proof countFun-in-01 (previous step)

i ≤ i0, i0 ≤ j, i ≤ j ⊦ intfun(i0) ∈ countFun(i, j, intfun)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ i0 ∈ countFun(i, j, intfun) ↔ (∃ j0. i ≤ j0 ∧ j0 ≤ j ∧ intfun(j0) = i0)
Proof step 4:
i ≤ i0, i0 ≤ j, i ≤ j ⊦ ∃ j0. i ≤ j0 ∧ j0 ≤ j ∧ intfun(j0) = intfun(i0)

Rule argument

spec: inst: name: countFun-in seq: ⊦ i0 ∈ countFun(i, j, intfun) ↔ (∃ j0. i ≤ j0 ∧ j0 ≤ j ∧ intfun(j0) = i0) subst: [intfun, j, i, i0] → [intfun, j, i, intfun(i0)] flags: rewrite, current

Additional information

Used terms: intfun, j, i, intfun(i0)