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Proof step 2 from proof map-in (previous step)

∀ i, intfun. i ∈ mapFun(intfun, ints0) ↔ (∃ j. j ∈ ints0 ∧ intfun(j) = i) ⊦ i ∈ mapFun(intfun, i0 + ints0) ↔ (∃ j. j ∈ i0 + ints0 ∧ intfun(j) = i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ i, intfun. i ∈ mapFun(intfun, ints0) ↔ (∃ j. j ∈ ints0 ∧ intfun(j) = i) ⊦ i = intfun(i0) ∨ i ∈ mapFun(intfun, ints0) ↔ (∃ j. (j = i0 ∨ j ∈ ints0) ∧ intfun(j) = i)

Rule argument

Additional information

Used simplifier rules: mapFun-rec: ⊦ mapFun(intfun, i ' + ints) = intfun(i) ' + mapFun(intfun, ints);%in-17: ⊦ i ∈ i0 ' ↔ i = i0; from specification list[intlista]%in-append: ⊦ i ∈ ints + ints0 ↔ ¬ ¬(i ∈ ints ∨ i ∈ ints0); from specification list[intlista]%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%c: ⊦ i + ints = i ' + ints; from specification list[intlista]