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Proof step 6 from proof map-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
intfun(i0) ∈ mapFun(intfun, ints0) ↔ (∃ j. j ∈ ints0 ∧ intfun(j) = intfun(i0)), ∀ i, intfun. i ∈ mapFun(intfun, ints0) ↔ (∃ j. j ∈ ints0 ∧ intfun(j) = i) ⊦ ∃ j. (j = i0 ∨ j ∈ ints0) ∧ intfun(j) = intfun(i0)

Proof step 6 from proof map-in (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information